Inorder Successor in Binary Search Tree

17 Jun, 2022
In Binary Tree, Inorder successor of a node is the next node in Inorder traversal of the Binary Tree. Inorder Successor is NULL for the last node in Inorder traversal. 
In Binary Search Tree, Inorder Successor of an input node can also be defined as the node with the smallest key greater than the key of the input node. So, it is sometimes important to find next node in sorted order.
In the above diagram, inorder successor of 8 is 10, inorder successor of 10 is 12 and inorder successor of 14 is 20.
Method 1 (Uses Parent Pointer) 
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In this method, we assume that every node has a parent pointer. The Algorithm is divided into two cases on the basis of the right subtree of the input node being empty or not.
Input: node, root // node is the node whose Inorder successor is needed. 
Output: succ // succ is Inorder successor of node.
If right subtree of node is not NULL, then succ lies in right subtree. Do the following. Go to right subtree and return the node with minimum key value in the right subtree.If right subtree of node is NULL, then succ is one of the ancestors. Do the following. Travel up using the parent pointer until you see a node which is left child of its parent. The parent of such a node is the succ.
If right subtree of node is not NULL, then succ lies in right subtree. Do the following. Go to right subtree and return the node with minimum key value in the right subtree.
If right subtree of node is NULL, then succ is one of the ancestors. Do the following. Travel up using the parent pointer until you see a node which is left child of its parent. The parent of such a node is the succ.
Implementation: 
Note that the function to find InOrder Successor is highlighted (with gray background) in below code.  
C++
C
Java
Python3
C#
Javascript
#include <iostream>using namespace std; /* A binary tree node has data,   the pointer to left child   and a pointer to right child */struct node {    int data;    struct node* left;    struct node* right;    struct node* parent;}; struct node* minValue(struct node* node); struct node* inOrderSuccessor(    struct node* root,    struct node* n){    // step 1 of the above algorithm    if (n->right != NULL)        return minValue(n->right);     // step 2 of the above algorithm    struct node* p = n->parent;    while (p != NULL && n == p->right) {        n = p;        p = p->parent;    }    return p;} /* Given a non-empty binary search tree,    return the minimum data     value found in that tree. Note that    the entire tree does not need    to be searched. */struct node* minValue(struct node* node){    struct node* current = node;     /* loop down to find the leftmost leaf */    while (current->left != NULL) {        current = current->left;    }    return current;} /* Helper function that allocates a new    node with the given data and    NULL left and right pointers. */struct node* newNode(int data){    struct node* node = (struct node*)        malloc(sizeof(            struct node));    node->data = data;    node->left = NULL;    node->right = NULL;    node->parent = NULL;     return (node);} /* Give a binary search tree and   a number, inserts a new node with       the given number in the correct    place in the tree. Returns the new    root pointer which the caller should    then use (the standard trick to    avoid using reference parameters). */struct node* insert(struct node* node,                    int data){    /* 1. If the tree is empty, return a new,      single node */    if (node == NULL)        return (newNode(data));    else {        struct node* temp;         /* 2. Otherwise, recur down the tree */        if (data <= node->data) {            temp = insert(node->left, data);            node->left = temp;            temp->parent = node;        }        else {            temp = insert(node->right, data);            node->right = temp;            temp->parent = node;        }         /* return the (unchanged) node pointer */        return node;    }} /* Driver program to test above functions*/int main(){    struct node *root = NULL, *temp, *succ, *min;     // creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root->left->right->right;     succ = inOrderSuccessor(root, temp);    if (succ != NULL)        cout << "\n Inorder Successor of " << temp->data<< " is "<< succ->data;    else        cout <<"\n Inorder Successor doesn't exit";     getchar();    return 0;} // this code is contributed by shivanisinghss2110
#include <stdio.h>#include <stdlib.h> /* A binary tree node has data,   the pointer to left child   and a pointer to right child */struct node {    int data;    struct node* left;    struct node* right;    struct node* parent;}; struct node* minValue(struct node* node); struct node* inOrderSuccessor(    struct node* root,    struct node* n){    // step 1 of the above algorithm    if (n->right != NULL)        return minValue(n->right);     // step 2 of the above algorithm    struct node* p = n->parent;    while (p != NULL && n == p->right) {        n = p;        p = p->parent;    }    return p;} /* Given a non-empty binary search tree,    return the minimum data     value found in that tree. Note that    the entire tree does not need    to be searched. */struct node* minValue(struct node* node){    struct node* current = node;     /* loop down to find the leftmost leaf */    while (current->left != NULL) {        current = current->left;    }    return current;} /* Helper function that allocates a new    node with the given data and    NULL left and right pointers. */struct node* newNode(int data){    struct node* node = (struct node*)        malloc(sizeof(            struct node));    node->data = data;    node->left = NULL;    node->right = NULL;    node->parent = NULL;     return (node);} /* Give a binary search tree and   a number, inserts a new node with       the given number in the correct    place in the tree. Returns the new    root pointer which the caller should    then use (the standard trick to    avoid using reference parameters). */struct node* insert(struct node* node,                    int data){    /* 1. If the tree is empty, return a new,      single node */    if (node == NULL)        return (newNode(data));    else {        struct node* temp;         /* 2. Otherwise, recur down the tree */        if (data <= node->data) {            temp = insert(node->left, data);            node->left = temp;            temp->parent = node;        }        else {            temp = insert(node->right, data);            node->right = temp;            temp->parent = node;        }         /* return the (unchanged) node pointer */        return node;    }} /* Driver program to test above functions*/int main(){    struct node *root = NULL, *temp, *succ, *min;     // creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root->left->right->right;     succ = inOrderSuccessor(root, temp);    if (succ != NULL)        printf(            "\n Inorder Successor of %d is %d ",            temp->data, succ->data);    else        printf("\n Inorder Successor doesn't exit");     getchar();    return 0;}
// Java program to find minimum// value node in Binary Search Tree // A binary tree nodeclass Node {     int data;    Node left, right, parent;     Node(int d)    {        data = d;        left = right = parent = null;    }} class BinaryTree {     static Node head;     /* Given a binary search tree and a number,     inserts a new node with the given number in     the correct place in the tree. Returns the new     root pointer which the caller should then use     (the standard trick to avoid using reference     parameters). */    Node insert(Node node, int data)    {         /* 1. If the tree is empty, return a new,             single node */        if (node == null) {            return (new Node(data));        }        else {             Node temp = null;             /* 2. Otherwise, recur down the tree */            if (data <= node.data) {                temp = insert(node.left, data);                node.left = temp;                temp.parent = node;            }            else {                temp = insert(node.right, data);                node.right = temp;                temp.parent = node;            }             /* return the (unchanged) node pointer */            return node;        }    }     Node inOrderSuccessor(Node root, Node n)    {         // step 1 of the above algorithm        if (n.right != null) {            return minValue(n.right);        }         // step 2 of the above algorithm        Node p = n.parent;        while (p != null && n == p.right) {            n = p;            p = p.parent;        }        return p;    }     /* Given a non-empty binary search       tree, return the minimum data        value found in that tree. Note that       the entire tree does not need       to be searched. */    Node minValue(Node node)    {        Node current = node;         /* loop down to find the leftmost leaf */        while (current.left != null) {            current = current.left;        }        return current;    }     // Driver program to test above functions    public static void main(String[] args)    {        BinaryTree tree = new BinaryTree();        Node root = null, temp = null, suc = null, min = null;        root = tree.insert(root, 20);        root = tree.insert(root, 8);        root = tree.insert(root, 22);        root = tree.insert(root, 4);        root = tree.insert(root, 12);        root = tree.insert(root, 10);        root = tree.insert(root, 14);        temp = root.left.right.right;        suc = tree.inOrderSuccessor(root, temp);        if (suc != null) {            System.out.println(                "Inorder successor of "                + temp.data + " is " + suc.data);        }        else {            System.out.println(                "Inorder successor does not exist");        }    }} // This code has been contributed by Mayank Jaiswal
# Python program to find the inorder successor in a BST # A binary tree nodeclass Node:     # Constructor to create a new node    def __init__(self, key):        self.data = key        self.left = None        self.right = None def inOrderSuccessor(n):         # Step 1 of the above algorithm    if n.right is not None:        return minValue(n.right)     # Step 2 of the above algorithm    p = n.parent    while( p is not None):        if n != p.right :            break        n = p        p = p.parent    return p # Given a non-empty binary search tree, return the# minimum data value found in that tree. Note that the# entire tree doesn't need to be searcheddef minValue(node):    current = node     # loop down to find the leftmost leaf    while(current is not None):        if current.left is None:            break        current = current.left     return current  # Given a binary search tree and a number, inserts a# new node with the given number in the correct place# in the tree. Returns the new root pointer which the# caller should then use( the standard trick to avoid# using reference parameters)def insert( node, data):     # 1) If tree is empty then return a new singly node    if node is None:        return Node(data)    else:                # 2) Otherwise, recur down the tree        if data <= node.data:            temp = insert(node.left, data)            node.left = temp            temp.parent = node        else:            temp = insert(node.right, data)            node.right = temp            temp.parent = node                 # return  the unchanged node pointer        return node  # Driver program to test above function root = None # Creating the tree given in the above diagramroot = insert(root, 20)root = insert(root, 8);root = insert(root, 22);root = insert(root, 4);root = insert(root, 12);root = insert(root, 10); root = insert(root, 14);   temp = root.left.right.right succ = inOrderSuccessor(temp)if succ is not None:    print ("\nInorder Successor of % d is % d "%(temp.data, succ.data))else:    print ("\nInorder Successor doesn't exist") # This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// C# program to find minimum// value node in Binary Search Treeusing System; // A binary tree nodepublic  class Node  {    public      int data;    public      Node left, right, parent;    public      Node(int d)    {      data = d;      left = right = parent = null;    }  } public class BinaryTree{  static Node head;   /* Given a binary search tree and a number,     inserts a new node with the given number in     the correct place in the tree. Returns the new     root pointer which the caller should then use     (the standard trick to avoid using reference     parameters). */  Node insert(Node node, int data)  {     /* 1. If the tree is empty, return a new,             single node */    if (node == null)    {      return (new Node(data));    }    else    {       Node temp = null;       /* 2. Otherwise, recur down the tree */      if (data <= node.data)      {        temp = insert(node.left, data);        node.left = temp;        temp.parent = node;      }      else      {        temp = insert(node.right, data);        node.right = temp;        temp.parent = node;      }       /* return the (unchanged) node pointer */      return node;    }  }   Node inOrderSuccessor(Node root, Node n)  {     // step 1 of the above algorithm    if (n.right != null)    {      return minValue(n.right);    }     // step 2 of the above algorithm    Node p = n.parent;    while (p != null && n == p.right)    {      n = p;      p = p.parent;    }    return p;  }   /* Given a non-empty binary search       tree, return the minimum data        value found in that tree. Note that       the entire tree does not need       to be searched. */  Node minValue(Node node)  {    Node current = node;     /* loop down to find the leftmost leaf */    while (current.left != null)    {      current = current.left;    }    return current;  }   // Driver program to test above functions  public static void Main(String[] args)  {    BinaryTree tree = new BinaryTree();    Node root = null, temp = null, suc = null, min = null;    root = tree.insert(root, 20);    root = tree.insert(root, 8);    root = tree.insert(root, 22);    root = tree.insert(root, 4);    root = tree.insert(root, 12);    root = tree.insert(root, 10);    root = tree.insert(root, 14);    temp = root.left.right.right;    suc = tree.inOrderSuccessor(root, temp);    if (suc != null) {      Console.WriteLine(        "Inorder successor of "        + temp.data + " is " + suc.data);    }    else {      Console.WriteLine(        "Inorder successor does not exist");    }  }} // This code is contributed by aashish1995
<script> // JavaScript program to find minimum// value node in Binary Search Tree // A binary tree nodeclass Node {    constructor(val) {        this.data = val;        this.left = null;        this.right = null;        this.parent = null;    }}      var head;     /*     * Given a binary search tree and a number,     inserts a new node with the given     * number in the correct place in the tree.     Returns the new root pointer which     * the caller should then use     (the standard trick to afunction using reference     * parameters).     */    function insert(node , data) {         /*         * 1. If the tree is empty,         return a new, single node         */        if (node == null) {            return (new Node(data));        } else {             var temp = null;             /* 2. Otherwise, recur down the tree */            if (data <= node.data) {                temp = insert(node.left, data);                node.left = temp;                temp.parent = node;            } else {                temp = insert(node.right, data);                node.right = temp;                temp.parent = node;            }             /* return the (unchanged) node pointer */            return node;        }    }     function inOrderSuccessor(root,  n) {         // step 1 of the above algorithm        if (n.right != null) {            return minValue(n.right);        }         // step 2 of the above algorithm        var p = n.parent;        while (p != null && n == p.right) {            n = p;            p = p.parent;        }        return p;    }     /*     * Given a non-empty binary search tree,     return the minimum data value found in     * that tree. Note that the entire tree     does not need to be searched.     */    function minValue(node) {        var current = node;         /* loop down to find the leftmost leaf */        while (current.left != null) {            current = current.left;        }        return current;    }     // Driver program to test above functions             var root = null, temp = null,        suc = null, min = null;        root = insert(root, 20);        root = insert(root, 8);        root = insert(root, 22);        root = insert(root, 4);        root = insert(root, 12);        root = insert(root, 10);        root = insert(root, 14);        temp = root.left.right.right;        suc = inOrderSuccessor(root, temp);        if (suc != null) {            document.write("Inorder successor of " +            temp.data + " is " + suc.data);        } else {            document.write(            "Inorder successor does not exist"            );        } // This code contributed by gauravrajput1 </script>
 Inorder Successor of 14 is 20
Complexity Analysis: 
Time Complexity: 
O(h), where h is the height of the tree. As in the second case(suppose skewed tree) we have to travel all the way towards the root.
Auxiliary Space: O(1). Due to no use of any data structure for storing values.
Method 2 (Search from root) Parent pointer is NOT needed in this algorithm. The Algorithm is divided into two cases on the basis of right subtree of the input node being empty or not.
Input: node, root // node is the node whose Inorder successor is needed. 
Output: succ // succ is Inorder successor of node.
If right subtree of node is not NULL, then succ lies in right subtree. Do the following. Go to right subtree and return the node with minimum key value in the right subtree.If right subtree of node is NULL, then start from the root and use search-like technique. Do the following. Travel down the tree, if a node’s data is greater than root’s data then go right side, otherwise, go to left side.
If right subtree of node is not NULL, then succ lies in right subtree. Do the following. Go to right subtree and return the node with minimum key value in the right subtree.
If right subtree of node is NULL, then start from the root and use search-like technique. Do the following. Travel down the tree, if a node’s data is greater than root’s data then go right side, otherwise, go to left side.
Below is the implementation of the above approach:
C++
C
Java
Python3
C#
Javascript
// C++ program for above approach#include <iostream>using namespace std; /* A binary tree node has data,   the pointer to left child   and a pointer to right child */struct node{    int data;    struct node* left;    struct node* right;    struct node* parent;}; struct node* minValue(struct node* node); struct node* inOrderSuccessor(struct node* root,                              struct node* n){         // Step 1 of the above algorithm    if (n->right != NULL)        return minValue(n->right);     struct node* succ = NULL;     // Start from root and search for    // successor down the tree    while (root != NULL)    {        if (n->data < root->data)        {            succ = root;            root = root->left;        }        else if (n->data > root->data)            root = root->right;        else            break;    }     return succ;} // Given a non-empty binary search tree,// return the minimum data value found// in that tree. Note that the entire// tree does not need to be searched.struct node* minValue(struct node* node){    struct node* current = node;     // Loop down to find the leftmost leaf    while (current->left != NULL)    {        current = current->left;    }    return current;} // Helper function that allocates a new// node with the given data and NULL left// and right pointers.struct node* newNode(int data){    struct node* node = (struct node*)    malloc(sizeof(struct node));    node->data = data;    node->left = NULL;    node->right = NULL;    node->parent = NULL;     return (node);} // Give a binary search tree and a// number, inserts a new node with// the given number in the correct// place in the tree. Returns the new// root pointer which the caller should// then use (the standard trick to// avoid using reference parameters).struct node* insert(struct node* node,                    int data){         /* 1. If the tree is empty, return a new,       single node */    if (node == NULL)        return (newNode(data));    else    {        struct node* temp;         /* 2. Otherwise, recur down the tree */        if (data <= node->data)        {            temp = insert(node->left, data);            node->left = temp;            temp->parent = node;        }        else        {            temp = insert(node->right, data);            node->right = temp;            temp->parent = node;        }         /* Return the (unchanged) node pointer */        return node;    }} // Driver codeint main(){    struct node *root = NULL, *temp, *succ, *min;     // Creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root->left->right->right;         // Function Call    succ = inOrderSuccessor(root, temp);    if (succ != NULL)        cout << "\n Inorder Successor of "             << temp->data << " is "<< succ->data;    else        cout <<"\n Inorder Successor doesn't exit";     getchar();    return 0;} // This code is contributed by shivanisinghss2110
// C program for above approach#include <stdio.h>#include <stdlib.h> /* A binary tree node has data,   the pointer to left child   and a pointer to right child */struct node{    int data;    struct node* left;    struct node* right;    struct node* parent;}; struct node* minValue(struct node* node); struct node* inOrderSuccessor(    struct node* root,    struct node* n){         // step 1 of the above algorithm    if (n->right != NULL)        return minValue(n->right);     struct node* succ = NULL;     // Start from root and search for    // successor down the tree    while (root != NULL)    {        if (n->data < root->data)        {            succ = root;            root = root->left;        }        else if (n->data > root->data)            root = root->right;        else            break;    }     return succ;} /* Given a non-empty binary search tree,    return the minimum data     value found in that tree. Note that    the entire tree does not need    to be searched. */struct node* minValue(struct node* node){    struct node* current = node;     /* loop down to find the leftmost leaf */    while (current->left != NULL)    {        current = current->left;    }    return current;} /* Helper function that allocates a new    node with the given data and    NULL left and right pointers. */struct node* newNode(int data){    struct node* node = (struct node*)        malloc(sizeof(            struct node));    node->data = data;    node->left = NULL;    node->right = NULL;    node->parent = NULL;     return (node);} /* Give a binary search tree and   a number, inserts a new node with       the given number in the correct    place in the tree. Returns the new    root pointer which the caller should    then use (the standard trick to    avoid using reference parameters). */struct node* insert(struct node* node,                    int data){    /* 1. If the tree is empty, return a new,      single node */    if (node == NULL)        return (newNode(data));    else    {        struct node* temp;         /* 2. Otherwise, recur down the tree */        if (data <= node->data)        {            temp = insert(node->left, data);            node->left = temp;            temp->parent = node;        }        else        {            temp = insert(node->right, data);            node->right = temp;            temp->parent = node;        }         /* return the (unchanged) node pointer */        return node;    }} /* Driver program to test above functions*/int main(){    struct node *root = NULL, *temp, *succ, *min;     // creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root->left->right->right;         // Function Call    succ = inOrderSuccessor(root, temp);    if (succ != NULL)        printf(            "\n Inorder Successor of %d is %d ",            temp->data, succ->data);    else        printf("\n Inorder Successor doesn't exit");     getchar();    return 0;} // Thanks to R.Srinivasan for suggesting this method.
// Java program for above approachclass GFG{   /* A binary tree node has data,   the pointer to left child   and a pointer to right child */static class node{    int data;    node left;    node right;    node parent;}; static node inOrderSuccessor(    node root,    node n){         // step 1 of the above algorithm    if (n.right != null)        return minValue(n.right);     node succ = null;     // Start from root and search for    // successor down the tree    while (root != null)    {        if (n.data < root.data)        {            succ = root;            root = root.left;        }        else if (n.data > root.data)            root = root.right;        else            break;    }    return succ;} /* Given a non-empty binary search tree,    return the minimum data     value found in that tree. Note that    the entire tree does not need    to be searched. */static node minValue(node node){    node current = node;     /* loop down to find the leftmost leaf */    while (current.left != null)    {        current = current.left;    }    return current;} /* Helper function that allocates a new    node with the given data and    null left and right pointers. */static node newNode(int data){    node node = new node();    node.data = data;    node.left = null;    node.right = null;    node.parent = null;     return (node);} /* Give a binary search tree and   a number, inserts a new node with       the given number in the correct    place in the tree. Returns the new    root pointer which the caller should    then use (the standard trick to    astatic void using reference parameters). */static  node insert(node node,                    int data){       /* 1. If the tree is empty, return a new,      single node */    if (node == null)        return (newNode(data));    else    {        node temp;         /* 2. Otherwise, recur down the tree */        if (data <= node.data)        {            temp = insert(node.left, data);            node.left = temp;            temp.parent = node;        }        else        {            temp = insert(node.right, data);            node.right = temp;            temp.parent = node;        }         /* return the (unchanged) node pointer */        return node;    }} /* Driver program to test above functions*/public static void main(String[] args){    node root = null, temp, succ, min;     // creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root.left.right.right;         // Function Call    succ = inOrderSuccessor(root, temp);    if (succ != null)        System.out.printf(            "\n Inorder Successor of %d is %d ",            temp.data, succ.data);    else        System.out.printf("\n Inorder Successor doesn't exit");}} // This code is contributed by gauravrajput1
# Python program to find# the inorder successor in a BST # A binary tree nodeclass Node:     # Constructor to create a new node    def __init__(self, key):        self.data = key        self.left = None        self.right = None def inOrderSuccessor(root, n):         # Step 1 of the above algorithm    if n.right is not None:        return minValue(n.right)     # Step 2 of the above algorithm    succ=Node(None)              while( root):        if(root.data<n.data):            root=root.right        elif(root.data>n.data):            succ=root            root=root.left        else:            break    return succ # Given a non-empty binary search tree,# return the minimum data value# found in that tree. Note that the# entire tree doesn't need to be searcheddef minValue(node):    current = node     # loop down to find the leftmost leaf    while(current is not None):        if current.left is None:            break        current = current.left     return current  # Given a binary search tree# and a number, inserts a# new node with the given# number in the correct place# in the tree. Returns the# new root pointer which the# caller should then use# (the standard trick to avoid# using reference parameters)def insert( node, data):     # 1) If tree is empty    # then return a new singly node    if node is None:        return Node(data)    else:                # 2) Otherwise, recur down the tree        if data <= node.data:            temp = insert(node.left, data)            node.left = temp            temp.parent = node        else:            temp = insert(node.right, data)            node.right = temp            temp.parent = node                 # return  the unchanged node pointer        return node  # Driver program to test above functionif __name__ == "__main__":  root = None   # Creating the tree given in the above diagram  root = insert(root, 20)  root = insert(root, 8);  root = insert(root, 22);  root = insert(root, 4);  root = insert(root, 12);  root = insert(root, 10);   root = insert(root, 14);     temp = root.left.right   succ = inOrderSuccessor( root, temp)  if succ is not None:      print("Inorder Successor of" ,               temp.data ,"is" ,succ.data)  else:      print("InInorder Successor doesn't exist")
// C# program for above approachusing System; public class GFG{   /* A binary tree node has data,   the pointer to left child   and a pointer to right child */  public    class node    {      public        int data;      public        node left;      public        node right;      public        node parent;    };   static node inOrderSuccessor(    node root,    node n)  {     // step 1 of the above algorithm    if (n.right != null)      return minValue(n.right);     node succ = null;     // Start from root and search for    // successor down the tree    while (root != null)    {      if (n.data < root.data)      {        succ = root;        root = root.left;      }      else if (n.data > root.data)        root = root.right;      else        break;    }    return succ;  }   /* Given a non-empty binary search tree,    return the minimum data     value found in that tree. Note that    the entire tree does not need    to be searched. */  static node minValue(node node)  {    node current = node;     /* loop down to find the leftmost leaf */    while (current.left != null)    {      current = current.left;    }    return current;  }   /* Helper function that allocates a new    node with the given data and    null left and right pointers. */  static node newNode(int data)  {    node node = new node();    node.data = data;    node.left = null;    node.right = null;    node.parent = null;     return (node);  }   /* Give a binary search tree and   a number, inserts a new node with       the given number in the correct    place in the tree. Returns the new    root pointer which the caller should    then use (the standard trick to    astatic void using reference parameters). */  static  node insert(node node,                      int data)  {     /* 1. If the tree is empty, return a new,      single node */    if (node == null)      return (newNode(data));    else    {      node temp;       /* 2. Otherwise, recur down the tree */      if (data <= node.data)      {        temp = insert(node.left, data);        node.left = temp;        temp.parent = node;      }      else      {        temp = insert(node.right, data);        node.right = temp;        temp.parent = node;      }       /* return the (unchanged) node pointer */      return node;    }  }   /* Driver program to test above functions*/  public static void Main(String[] args)  {    node root = null, temp, succ;     // creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root.left.right.right;     // Function Call    succ = inOrderSuccessor(root, temp);    if (succ != null)      Console.Write(      "\n Inorder Successor of {0} is {1} ",      temp.data, succ.data);    else      Console.Write("\n Inorder Successor doesn't exit");  }} // This code is contributed by gauravrajput1
<script> class Node{    constructor(data)    {        this.data=data;;        this.left=this.right=this.parent=null;    }} function inOrderSuccessor(root,n){    // step 1 of the above algorithm    if (n.right != null)        return minValue(n.right);      let succ = null;      // Start from root and search for    // successor down the tree    while (root != null)    {        if (n.data < root.data)        {            succ = root;            root = root.left;        }        else if (n.data > root.data)            root = root.right;        else            break;    }    return succ;} function minValue(node){    let current = node;      /* loop down to find the leftmost leaf */    while (current.left != null)    {        current = current.left;    }    return current;}  function insert(node,data){    /* 1. If the tree is empty, return a new,      single node */    if (node == null)        return (new Node(data));    else    {        let temp;          /* 2. Otherwise, recur down the tree */        if (data <= node.data)        {            temp = insert(node.left, data);            node.left = temp;            temp.parent = node;        }        else        {            temp = insert(node.right, data);            node.right = temp;            temp.parent = node;        }          /* return the (unchanged) node pointer */        return node;    }} let root = null, temp, succ, min;  // creating the tree given in the above diagramroot = insert(root, 20);root = insert(root, 8);root = insert(root, 22);root = insert(root, 4);root = insert(root, 12);root = insert(root, 10);root = insert(root, 14);temp = root.left.right.right; // Function Callsucc = inOrderSuccessor(root, temp);if (succ != null)    document.write("<br> Inorder Successor of "+temp.data+"  is "+ succ.data);else    document.write("<br> Inorder Successor doesn't exit"); // This code is contributed by unknown2108 </script>
 Inorder Successor of 14 is 20
Complexity Analysis: 
Time Complexity: O(h), where h is the height of the tree. In the worst case as explained above we travel the whole height of the tree
Auxiliary Space: O(1). Due to no use of any data structure for storing values.
Method 3 (Inorder traversal) An inorder transversal of BST produces a sorted sequence. Therefore, we perform an inorder traversal. The first encountered node with value greater than the node is the inorder successor. 
Input: node, root // node is the node whose ignorer successor is needed.
Output: succ // succ is Inorder successor of node.
Below is the implementation of the above approach:
C++
Java
C#
Javascript
// C++ program for above approach#include <iostream>using namespace std; /* A binary tree node has data,   the pointer to left child   and a pointer to right child */struct node{    int data;    struct node* left;    struct node* right;    struct node* parent;};struct node* newNode(int data); void inOrderTraversal(struct node* root,                              struct node* n,                              struct node* succ){       if(root==nullptr) { return; }             inOrderTraversal(root->left, n, succ);    if(root->data>n->data && !succ->left) { succ->left = root; return; }    inOrderTraversal(root->right, n, succ);    } struct node* inOrderSuccessor(struct node* root,                              struct node* n){       struct node* succ = newNode(0);    inOrderTraversal(root, n, succ);    return succ->left;} // Helper function that allocates a new// node with the given data and NULL left// and right pointers.struct node* newNode(int data){    struct node* node = (struct node*)    malloc(sizeof(struct node));    node->data = data;    node->left = NULL;    node->right = NULL;    node->parent = NULL;     return (node);} // Give a binary search tree and a// number, inserts a new node with// the given number in the correct// place in the tree. Returns the new// root pointer which the caller should// then use (the standard trick to// avoid using reference parameters).struct node* insert(struct node* node,                    int data){         /* 1. If the tree is empty, return a new,       single node */    if (node == NULL)        return (newNode(data));    else    {        struct node* temp;         /* 2. Otherwise, recur down the tree */        if (data <= node->data)        {            temp = insert(node->left, data);            node->left = temp;            temp->parent = node;        }        else        {            temp = insert(node->right, data);            node->right = temp;            temp->parent = node;        }         /* Return the (unchanged) node pointer */        return node;    }} // Driver codeint main(){    struct node *root = NULL, *temp, *succ, *min;     // Creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root->left->right->right;         // Function Call    succ = inOrderSuccessor(root, temp);    if (succ != NULL)        cout << "\n Inorder Successor of "             << temp->data << " is "<< succ->data;    else        cout <<"\n Inorder Successor doesn't exist";     //getchar();    return 0;} // This code is contributed by jaisw7
// Java program for above approachimport java.util.*; class GFG {   /*     * A binary tree node has data, the pointer to left child and a pointer to right     * child     */  static class node {    int data;    node left;    node right;    node parent;  };  static void inOrderTraversal(node root) {    if (root == null) {      return;    }     inOrderTraversal(root.left);    System.out.print(root.data);    inOrderTraversal(root.right);  }  static void inOrderTraversal(node root, node n, node succ) {    if (root == null) {      return;    }     inOrderTraversal(root.left, n, succ);    if (root.data > n.data && succ.left == null) {      succ.left = root;      return;    }    inOrderTraversal(root.right, n, succ);  }   static node inOrderSuccessor(node root, node n) {    node succ = newNode(0);    inOrderTraversal(root, n, succ);    return succ.left;  }   // Helper function that allocates a new  // node with the given data and null left  // and right pointers.  static node newNode(int data) {    node node = new node();     node.data = data;    node.left = null;    node.right = null;    node.parent = null;     return (node);  }   // Give a binary search tree and a  // number, inserts a new node with  // the given number in the correct  // place in the tree. Returns the new  // root pointer which the caller should  // then use (the standard trick to  // astatic void using reference parameters).  static node insert(node node, int data) {     /*         * 1. If the tree is empty, return a new, single node         */    if (node == null)      return (newNode(data));    else {      node temp;       /* 2. Otherwise, recur down the tree */      if (data <= node.data) {        temp = insert(node.left, data);        node.left = temp;        temp.parent = node;      } else {        temp = insert(node.right, data);        node.right = temp;        temp.parent = node;      }       /* Return the (unchanged) node pointer */      return node;    }  }   // Driver code  public static void main(String[] args) {    node root = null, temp, succ, min;     // Creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root.left.right.right;     // Function Call    succ = inOrderSuccessor(root, temp);    if (succ != null)      System.out.print("\n Inorder Successor of " + temp.data + " is " + succ.data);    else      System.out.print("\n Inorder Successor doesn't exist");   }} // This code is contributed by Rajput-Ji
// C# program for above approachusing System;using System.Collections.Generic; public class GFG {   /*     * A binary tree node has data, the pointer to left child and a pointer to right     * child     */ public  class node {   public  int data;   public  node left;   public  node right;   public  node parent;  };  static void inOrderTraversal(node root) {    if (root == null) {      return;    }     inOrderTraversal(root.left);    Console.Write(root.data);    inOrderTraversal(root.right);  }  static void inOrderTraversal(node root, node n, node succ) {    if (root == null) {      return;    }     inOrderTraversal(root.left, n, succ);    if (root.data > n.data && succ.left == null) {      succ.left = root;      return;    }    inOrderTraversal(root.right, n, succ);  }   static node inOrderSuccessor(node root, node n) {    node succ = newNode(0);    inOrderTraversal(root, n, succ);    return succ.left;  }   // Helper function that allocates a new  // node with the given data and null left  // and right pointers.  static node newNode(int data) {    node node = new node();     node.data = data;    node.left = null;    node.right = null;    node.parent = null;     return (node);  }   // Give a binary search tree and a  // number, inserts a new node with  // the given number in the correct  // place in the tree. Returns the new  // root pointer which the caller should  // then use (the standard trick to  // astatic void using reference parameters).  static node insert(node node, int data) {     /*         * 1. If the tree is empty, return a new, single node         */    if (node == null)      return (newNode(data));    else {      node temp;       /* 2. Otherwise, recur down the tree */      if (data <= node.data) {        temp = insert(node.left, data);        node.left = temp;        temp.parent = node;      } else {        temp = insert(node.right, data);        node.right = temp;        temp.parent = node;      }       /* Return the (unchanged) node pointer */      return node;    }  }   // Driver code  public static void Main(String[] args) {    node root = null, temp, succ, min;     // Creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root.left.right.right;     // Function Call    succ = inOrderSuccessor(root, temp);    if (succ != null)      Console.Write("\n Inorder Successor of " + temp.data + " is " + succ.data);    else      Console.Write("\n Inorder Successor doesn't exist");   }} // This code contributed by Rajput-Ji
<script>// javascript program for above approach     /*     * A binary tree node has data, the pointer to left child and a pointer to right     * child     */     class Node {         constructor(){        this.data = 0;        this.left = null;        this.right = null;        this.parent = null;    }    }     function inOrderTraversal( root) {        if (root == null) {            return;        }         inOrderTraversal(root.left);        document.write(root.data);        inOrderTraversal(root.right);    }     function inOrderTraversal( root,  n,  succ) {        if (root == null) {            return;        }         inOrderTraversal(root.left, n, succ);        if (root.data > n.data && succ.left == null) {            succ.left = root;            return;        }        inOrderTraversal(root.right, n, succ);    }      function inOrderSuccessor( root,  n) {        var succ = newNode(0);        inOrderTraversal(root, n, succ);        return succ.left;    }     // Helper function that allocates a new    // node with the given data and null left    // and right pointers.     function newNode(data) {        var node = new Node();         node.data = data;        node.left = null;        node.right = null;        node.parent = null;         return (node);    }     // Give a binary search tree and a    // number, inserts a new node with    // the given number in the correct    // place in the tree. Returns the new    // root pointer which the caller should    // then use (the standard trick to    // afunction using reference parameters).     function insert( node , data) {         /*         * 1. If the tree is empty, return a new, single node         */        if (node == null)            return (newNode(data));        else {            var temp;             /* 2. Otherwise, recur down the tree */            if (data <= node.data) {                temp = insert(node.left, data);                node.left = temp;                temp.parent = node;            } else {                temp = insert(node.right, data);                node.right = temp;                temp.parent = node;            }             /* Return the (unchanged) node pointer */            return node;        }    }     // Driver code             var root = null, temp, succ, min;         // Creating the tree given in the above diagram        root = insert(root, 20);        root = insert(root, 8);        root = insert(root, 22);        root = insert(root, 4);        root = insert(root, 12);        root = insert(root, 10);        root = insert(root, 14);        temp = root.left.right.right;         // Function Call        succ = inOrderSuccessor(root, temp);        if (succ != null)            document.write("\n Inorder Successor of " + temp.data +            " is " + succ.data);        else            document.write("\n Inorder Successor doesn't exist"); // This code contributed by Rajput-Ji</script>
 Inorder Successor of 14 is 20
Complexity Analysis:
Time Complexity: O(h), where h is the height of the tree. In the worst case as explained above we travel the whole height of the tree.
Auxiliary Space: O(1). Due to no use of any data structure for storing values.
Method 4 (Inorder traversal iterative) this method is inspired from the method 3 but with iterative and easy to understand approach.
Input: node, root // node is the node whose inorder successor is needed.Output: succ // succ is Inorder successor of node.
Below is the implementation of the above approach:
Java
// Java program for above approachimport java.util.*; class GFG {   /*     * A binary tree node has data, the pointer to left child and a pointer to right     * child     */  static class node {    int data;    node left;    node right;    node parent;  };  static void inOrderTraversal(node root) {    if (root == null) {      return;    }     inOrderTraversal(root.left);    System.out.print(root.data);    inOrderTraversal(root.right);  }    public static node inOrderSuccessor(node root, int key) {        Deque<node> stack = new ArrayDeque<>();        while(root != null || !stack.isEmpty()){            while(root != null){                stack.push(root);                root = root.left;            }            root = stack.pop();            if(root.data > key)                return root;            root = root.right;        }        return null;    }   // Helper function that allocates a new  // node with the given data and null left  // and right pointers.  static node newNode(int data) {    node node = new node();     node.data = data;    node.left = null;    node.right = null;    node.parent = null;     return (node);  }   // Give a binary search tree and a  // number, inserts a new node with  // the given number in the correct  // place in the tree. Returns the new  // root pointer which the caller should  // then use (the standard trick to  // astatic void using reference parameters).  static node insert(node node, int data) {     /*         * 1. If the tree is empty, return a new, single node         */    if (node == null)      return (newNode(data));    else {      node temp;       /* 2. Otherwise, recur down the tree */      if (data <= node.data) {        temp = insert(node.left, data);        node.left = temp;        temp.parent = node;      } else {        temp = insert(node.right, data);        node.right = temp;        temp.parent = node;      }       /* Return the (unchanged) node pointer */      return node;    }  }   // Driver code  public static void main(String[] args) {    node root = null, temp, succ, min;     // Creating the tree given in the above diagram    root = insert(root, 20);    root = insert(root, 8);    root = insert(root, 22);    root = insert(root, 4);    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 14);    temp = root.left.right.right;     // Function Call    succ = inOrderSuccessor(root, temp.data);    if (succ != null)      System.out.print("\n Inorder Successor of " + temp.data + " is " + succ.data);    else      System.out.print("\n Inorder Successor doesn't exist");   }} // This code is contributed by Nitin Dhamija
 Inorder Successor of 14 is 20
Complexity Analysis:
Time Complexity: O(h), where h is the height of the tree. In the worst case as explained above we travel the whole height of the tree
Auxiliary Space: O(1). Due to no use of any data structure for storing values.
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Binary Search Tree
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Binary Search Tree
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