Convert a given Binary Tree to Doubly Linked List | Set 1 - GeeksforGeeks

28 Jan, 2022
Given a Binary Tree (Bt), convert it to a Doubly Linked List(DLL). The left and right pointers in nodes are to be used as previous and next pointers respectively in converted DLL. The order of nodes in DLL must be the same as in Inorder for the given Binary Tree. The first node of Inorder traversal (leftmost node in BT) must be the head node of the DLL. 
I came across this question during one of my interviews. A similar problem has been discussed in this post. 
The problem here is simpler as we don’t need to create a circular DLL, but a simple DLL. The idea behind its solution is quite simple and straight.
If the left subtree exists, process the left subtreeRecursively convert the left subtree to DLL.Then find the inorder predecessor of the root in the left subtree (the inorder predecessor is the rightmost node in the left subtree).Make the inorder predecessor as the previous root and the root as the next in order predecessor. If the right subtree exists, process the right subtree (Below 3 steps are similar to the left subtree).Recursively convert the right subtree to DLL.Then find the inorder successor of the root in the right subtree (in order the successor is the leftmost node in the right subtree).Make the inorder successor as the next root and the root as the previous inorder successor.Find the leftmost node and return it (the leftmost node is always the head of a converted DLL).
If the left subtree exists, process the left subtreeRecursively convert the left subtree to DLL.Then find the inorder predecessor of the root in the left subtree (the inorder predecessor is the rightmost node in the left subtree).Make the inorder predecessor as the previous root and the root as the next in order predecessor.
Recursively convert the left subtree to DLL.Then find the inorder predecessor of the root in the left subtree (the inorder predecessor is the rightmost node in the left subtree).Make the inorder predecessor as the previous root and the root as the next in order predecessor.
Recursively convert the left subtree to DLL.
Then find the inorder predecessor of the root in the left subtree (the inorder predecessor is the rightmost node in the left subtree).
Make the inorder predecessor as the previous root and the root as the next in order predecessor.
 If the right subtree exists, process the right subtree (Below 3 steps are similar to the left subtree).Recursively convert the right subtree to DLL.Then find the inorder successor of the root in the right subtree (in order the successor is the leftmost node in the right subtree).Make the inorder successor as the next root and the root as the previous inorder successor.
Recursively convert the right subtree to DLL.Then find the inorder successor of the root in the right subtree (in order the successor is the leftmost node in the right subtree).Make the inorder successor as the next root and the root as the previous inorder successor.
Recursively convert the right subtree to DLL.
Then find the inorder successor of the root in the right subtree (in order the successor is the leftmost node in the right subtree).
Make the inorder successor as the next root and the root as the previous inorder successor.
Find the leftmost node and return it (the leftmost node is always the head of a converted DLL).
Below is the source code for the above algorithm.
C++
C
Java
Python3
C#
Javascript
// A C++ program for in-place// conversion of Binary Tree to DLL#include <bits/stdc++.h>using namespace std; /* A binary tree node has data,and left and right pointers */class node {public:    int data;    node* left;    node* right;}; /* This is the core function to convertTree to list. This function followssteps 1 and 2 of the above algorithm */node* bintree2listUtil(node* root){    // Base case    if (root == NULL)        return root;     // Convert the left subtree and link to root    if (root->left != NULL) {        // Convert the left subtree        node* left = bintree2listUtil(root->left);         // Find inorder predecessor. After this loop, left        // will point to the inorder predecessor        for (; left->right != NULL; left = left->right)            ;         // Make root as next of the predecessor        left->right = root;         // Make predecessor as previous of root        root->left = left;    }     // Convert the right subtree and link to root    if (root->right != NULL) {        // Convert the right subtree        node* right = bintree2listUtil(root->right);         // Find inorder successor. After this loop, right        // will point to the inorder successor        for (; right->left != NULL; right = right->left)            ;         // Make root as previous of successor        right->left = root;         // Make successor as next of root        root->right = right;    }     return root;} // The main function that first calls// bintree2listUtil(), then follows step 3// of the above algorithmnode* bintree2list(node* root){    // Base case    if (root == NULL)        return root;     // Convert to DLL using bintree2listUtil()    root = bintree2listUtil(root);     // bintree2listUtil() returns root node of the converted    // DLL. We need pointer to the leftmost node which is    // head of the constructed DLL, so move to the leftmost    // node    while (root->left != NULL)        root = root->left;     return (root);} /* Helper function that allocates a new node with thegiven data and NULL left and right pointers. */node* newNode(int data){    node* new_node = new node();    new_node->data = data;    new_node->left = new_node->right = NULL;    return (new_node);} /* Function to print nodes in a given doubly linked list */void printList(node* node){    while (node != NULL) {        cout << node->data << " ";        node = node->right;    }} /* Driver code*/int main(){    // Let us create the tree shown in above diagram    node* root = newNode(10);    root->left = newNode(12);    root->right = newNode(15);    root->left->left = newNode(25);    root->left->right = newNode(30);    root->right->left = newNode(36);     // Convert to DLL    node* head = bintree2list(root);     // Print the converted list    printList(head);     return 0;} // This code is contributed by rathbhupendra
// A C program for in-place conversion of Binary Tree to DLL#include <stdio.h> /* A binary tree node has data, and left and right pointers */struct node{    int data;    node* left;    node* right;}; /* This is the core function to convert Tree to list. This function follows  steps 1 and 2 of the above algorithm */node* bintree2listUtil(node* root){    // Base case    if (root == NULL)        return root;     // Convert the left subtree and link to root    if (root->left != NULL)    {        // Convert the left subtree        node* left = bintree2listUtil(root->left);         // Find inorder predecessor. After this loop, left        // will point to the inorder predecessor        for (; left->right!=NULL; left=left->right);         // Make root as next of the predecessor        left->right = root;         // Make predecessor as previous of root        root->left = left;    }     // Convert the right subtree and link to root    if (root->right!=NULL)    {        // Convert the right subtree        node* right = bintree2listUtil(root->right);         // Find inorder successor. After this loop, right        // will point to the inorder successor        for (; right->left!=NULL; right = right->left);         // Make root as previous of successor        right->left = root;         // Make successor as next of root        root->right = right;    }     return root;} // The main function that first calls bintree2listUtil(), then follows step 3//  of the above algorithmnode* bintree2list(node *root){    // Base case    if (root == NULL)        return root;     // Convert to DLL using bintree2listUtil()    root = bintree2listUtil(root);     // bintree2listUtil() returns root node of the converted    // DLL.  We need pointer to the leftmost node which is    // head of the constructed DLL, so move to the leftmost node    while (root->left != NULL)        root = root->left;     return (root);} /* Helper function that allocates a new node with the   given data and NULL left and right pointers. */node* newNode(int data){    node* new_node = new node;    new_node->data = data;    new_node->left = new_node->right = NULL;    return (new_node);} /* Function to print nodes in a given doubly linked list */void printList(node *node){    while (node!=NULL)    {        printf("%d ", node->data);        node = node->right;    }} /* Driver program to test above functions*/int main(){    // Let us create the tree shown in above diagram    node *root        = newNode(10);    root->left        = newNode(12);    root->right       = newNode(15);    root->left->left  = newNode(25);    root->left->right = newNode(30);    root->right->left = newNode(36);     // Convert to DLL    node *head = bintree2list(root);     // Print the converted list    printList(head);     return 0;}
// Java program to convert binary tree to double linked list  /* A binary tree node has data, and left and right pointers */class Node{    int data;    Node left, right;      Node(int item)    {        data = item;        left = right = null;    }}  class BinaryTree{    Node root;    /* This is the core function to convert Tree to list. This function       follows steps 1 and 2 of the above algorithm */      Node bintree2listUtil(Node node)    {        // Base case        if (node == null)            return node;          // Convert the left subtree and link to root        if (node.left != null)        {            // Convert the left subtree            Node left = bintree2listUtil(node.left);              // Find inorder predecessor. After this loop, left            // will point to the inorder predecessor            for (; left.right != null; left = left.right);              // Make root as next of the predecessor            left.right = node;              // Make predecessor as previous of root            node.left = left;        }          // Convert the right subtree and link to root        if (node.right != null)        {            // Convert the right subtree            Node right = bintree2listUtil(node.right);              // Find inorder successor. After this loop, right            // will point to the inorder successor            for (; right.left != null; right = right.left);              // Make root as previous of successor            right.left = node;              // Make successor as next of root            node.right = right;        }          return node;    }      // The main function that first calls bintree2listUtil(), then follows    // step 3 of the above algorithm          Node bintree2list(Node node)    {        // Base case        if (node == null)            return node;          // Convert to DLL using bintree2listUtil()        node = bintree2listUtil(node);          // bintree2listUtil() returns root node of the converted        // DLL.  We need pointer to the leftmost node which is        // head of the constructed DLL, so move to the leftmost node        while (node.left != null)            node = node.left;          return node;    }      /* Function to print nodes in a given doubly linked list */    void printList(Node node)    {        while (node != null)        {            System.out.print(node.data + " ");            node = node.right;        }    }      /* Driver program to test above functions*/    public static void main(String[] args)    {        BinaryTree tree = new BinaryTree();          // Let us create the tree shown in above diagram        tree.root = new Node(10);        tree.root.left = new Node(12);        tree.root.right = new Node(15);        tree.root.left.left = new Node(25);        tree.root.left.right = new Node(30);        tree.root.right.left = new Node(36);          // Convert to DLL        Node head = tree.bintree2list(tree.root);          // Print the converted list        tree.printList(head);    }}
# Python program to convert# binary tree to doubly linked list class Node(object):         """Binary tree Node class has    data, left and right child"""    def __init__(self, item):        self.data = item        self.left = None        self.right = None def BTToDLLUtil(root):         """This is a utility function to    convert the binary tree to doubly    linked list. Most of the core task    is done by this function."""    if root is None:        return root     # Convert left subtree    # and link to root    if root.left:                 # Convert the left subtree        left = BTToDLLUtil(root.left)         # Find inorder predecessor, After        # this loop, left will point to the        # inorder predecessor of root        while left.right:            left = left.right         # Make root as next of predecessor        left.right = root                 # Make predecessor as        # previous of root        root.left = left     # Convert the right subtree    # and link to root    if root.right:                 # Convert the right subtree        right = BTToDLLUtil(root.right)         # Find inorder successor, After        # this loop, right will point to        # the inorder successor of root        while right.left:            right = right.left         # Make root as previous        # of successor        right.left = root                 # Make successor as        # next of root        root.right = right     return root def BTToDLL(root):    if root is None:        return root     # Convert to doubly linked    # list using BLLToDLLUtil    root = BTToDLLUtil(root)         # We need pointer to left most    # node which is head of the    # constructed Doubly Linked list    while root.left:        root = root.left     return root def print_list(head):         """Function to print the given       doubly linked list"""    if head is None:        return    while head:        print(head.data, end = " ")        head = head.right # Driver Codeif __name__ == '__main__':    root = Node(10)    root.left = Node(12)    root.right = Node(15)    root.left.left = Node(25)    root.left.right = Node(30)    root.right.left = Node(36)     head = BTToDLL(root)    print_list(head) # This code is contributed# by viveksyngh
using System; // C# program to convert binary tree to double linked list /* A binary tree node has data, and left and right pointers */public class Node{    public int data;    public Node left, right;     public Node(int item)    {        data = item;        left = right = null;    }} public class BinaryTree{    public Node root;    /* This is the core function to convert Tree to list. This function       follows steps 1 and 2 of the above algorithm */     public virtual Node bintree2listUtil(Node node)    {        // Base case        if (node == null)        {            return node;        }         // Convert the left subtree and link to root        if (node.left != null)        {            // Convert the left subtree            Node left = bintree2listUtil(node.left);             // Find inorder predecessor. After this loop, left            // will point to the inorder predecessor            for (; left.right != null; left = left.right)            {                ;            }             // Make root as next of the predecessor            left.right = node;             // Make predecessor as previous of root            node.left = left;        }         // Convert the right subtree and link to root        if (node.right != null)        {            // Convert the right subtree            Node right = bintree2listUtil(node.right);             // Find inorder successor. After this loop, right            // will point to the inorder successor            for (; right.left != null; right = right.left)            {                ;            }             // Make root as previous of successor            right.left = node;             // Make successor as next of root            node.right = right;        }         return node;    }     // The main function that first calls bintree2listUtil(), then follows    // step 3 of the above algorithm     public virtual Node bintree2list(Node node)    {        // Base case        if (node == null)        {            return node;        }         // Convert to DLL using bintree2listUtil()        node = bintree2listUtil(node);         // bintree2listUtil() returns root node of the converted        // DLL.  We need pointer to the leftmost node which is        // head of the constructed DLL, so move to the leftmost node        while (node.left != null)        {            node = node.left;        }         return node;    }     /* Function to print nodes in a given doubly linked list */    public virtual void printList(Node node)    {        while (node != null)        {            Console.Write(node.data + " ");            node = node.right;        }    }     /* Driver program to test above functions*/    public static void Main(string[] args)    {        BinaryTree tree = new BinaryTree();         // Let us create the tree shown in above diagram        tree.root = new Node(10);        tree.root.left = new Node(12);        tree.root.right = new Node(15);        tree.root.left.left = new Node(25);        tree.root.left.right = new Node(30);        tree.root.right.left = new Node(36);         // Convert to DLL        Node head = tree.bintree2list(tree.root);         // Print the converted list        tree.printList(head);    }} // This code is contributed by Shrikant13
<script>    // javascript program to convert binary tree to double linked list     /* A binary tree node has data, and left and right pointers */    class Node {        constructor(val) {            this.data = val;            this.left = null;            this.right = null;        }    }    var root;    /*     * This is the core function to convert Tree to list. This function follows     * steps 1 and 2 of the above algorithm     */     function bintree2listUtil(node) {        // Base case        if (node == null)            return node;         // Convert the left subtree and link to root        if (node.left != null) {            // Convert the left subtree            var left = bintree2listUtil(node.left);             // Find inorder predecessor. After this loop, left            // will point to the inorder predecessor            for (; left.right != null; left = left.right)                               // Make root as next of the predecessor            left.right = node;             // Make predecessor as previous of root            node.left = left;        }         // Convert the right subtree and link to root        if (node.right != null) {            // Convert the right subtree            var right = bintree2listUtil(node.right);             // Find inorder successor. After this loop, right            // will point to the inorder successor            for (; right.left != null; right = right.left)                              // Make root as previous of successor            right.left = node;             // Make successor as next of root            node.right = right;        }         return node;    }     // The main function that first calls bintree2listUtil(), then follows    // step 3 of the above algorithm     function bintree2list(node) {        // Base case        if (node == null)            return node;         // Convert to DLL using bintree2listUtil()        node = bintree2listUtil(node);         // bintree2listUtil() returns root node of the converted        // DLL. We need pointer to the leftmost node which is        // head of the constructed DLL, so move to the leftmost node        while (node.left != null)            node = node.left;         return node;    }     /* Function to print nodes in a given doubly linked list */    function printList(node) {        while (node != null) {            document.write(node.data + " ");            node = node.right;        }    }     /* Driver program to test above functions */         // Let us create the tree shown in above diagram    root = new Node(10);    root.left = new Node(12);    root.right = new Node(15);    root.left.left = new Node(25);    root.left.right = new Node(30);    root.right.left = new Node(36);     // Convert to DLL    var head = bintree2list(root);     // Print the converted list    printList(head); // This code contributed by umadevi9616</script>
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Another Approach:Algorithm:
Traverse the tree in inorder fashion.While visiting each node, keep track of DLL’s head and tail pointers, insert each visited node to the end of DLL using tail pointer.Return head of the list.
Traverse the tree in inorder fashion.
While visiting each node, keep track of DLL’s head and tail pointers, insert each visited node to the end of DLL using tail pointer.
Return head of the list.
Below is the implementation of the above approach:
C++
Java
C#
Javascript
// A C++ program for in-place// conversion of Binary Tree to DLL#include <bits/stdc++.h>using namespace std; /* A binary tree node has data,and left and right pointers */class node {public:    int data;    node* left;    node* right;}; /* This is the core function to convertTree to list.*/void bintree2listUtil(node* root, node** head, node** tail){    if (root == NULL)        return;     bintree2listUtil(root->left, head, tail);     if (*head == NULL) {        *head = root;    }     root->left = *tail;     if (*tail != NULL) {        (*tail)->right = root;    }     *tail = root;     bintree2listUtil(root->right, head, tail);} // The main function that first calls// bintree2listUtil()node* bintree2list(node* root){    // Base case    if (root == NULL)        return root;     node* head = NULL;    node* tail = NULL;     bintree2listUtil(root, &head, &tail);     return head;} /* Helper function that allocates a new node with thegiven data and NULL left and right pointers. */node* newNode(int data){    node* new_node = new node();    new_node->data = data;    new_node->left = new_node->right = NULL;    return (new_node);} /* Function to print nodes in a given doubly linked list */void printList(node* node){    while (node != NULL) {        cout << node->data << " ";        node = node->right;    }} /* Driver code*/int main(){    // Let us create the tree shown in above diagram    node* root = newNode(10);    root->left = newNode(12);    root->right = newNode(15);    root->left->left = newNode(25);    root->left->right = newNode(30);    root->right->left = newNode(36);     // Convert to DLL    node* head = bintree2list(root);     // Print the converted list    printList(head);     return 0;} // This code is contributed by rathbhupendra
// A Java program for in-place// conversion of Binary Tree to DLLimport java.util.*; class GFG{ /* A binary tree node has data,and left and right pointers */static class node {     int data;    node left;    node right;};     /*     * This is the core function to convert Tree to list.     */static node head, tail;static void bintree2listUtil(node root){    if (root == null)        return ;     node left = root.left;    node right = root.right;     bintree2listUtil(root.left);     if (head == null) {        head = root;    }     root.left = tail;     if (tail != null) {        (tail).right = root;    }     tail = root;     bintree2listUtil(root.right);} // The main function that first calls// bintree2listUtil()static void bintree2list(node root){       // Base case    if (root == null)        head= root;         bintree2listUtil(root);     } /* Helper function that allocates a new node with thegiven data and null left and right pointers. */static node newNode(int data){    node new_node = new node();    new_node.data = data;    new_node.left = new_node.right = null;    return (new_node);}     /* Function to print nodes in a given doubly linked list */static void printList(){    while (head != null) {        System.out.print(head.data+ " ");        head = head.right;    }}     /* Driver code */public static void main(String[] args){       // Let us create the tree shown in above diagram    node root = newNode(10);    root.left = newNode(12);    root.right = newNode(15);    root.left.left = newNode(25);    root.left.right = newNode(30);    root.right.left = newNode(36);    head = null;    tail = null;       // Convert to DLL    bintree2list(root);     // Print the converted list    printList(); }} // This code is contributed by umadevi9616
// A C# program for in-place// conversion of Binary Tree to DLLusing System; public class GFG {     /*     * A binary tree node has data, and left and right pointers     */public    class node {     public    int data;    public    node left;    public    node right;    };     /*     * This is the core function to convert Tree to list.     */    static node head, tail;     static void bintree2listUtil(node root) {        if (root == null)            return;         node left = root.left;        node right = root.right;         bintree2listUtil(root.left);         if (head == null) {            head = root;        }         root.left = tail;         if (tail != null) {            (tail).right = root;        }         tail = root;         bintree2listUtil(root.right);    }     // The main function that first calls    // bintree2listUtil()    static void bintree2list(node root) {         // Base case        if (root == null)            head = root;         bintree2listUtil(root);     }     /*     * Helper function that allocates a new node with the given data and null left     * and right pointers.     */    static node newNode(int data) {        node new_node = new node();        new_node.data = data;        new_node.left = new_node.right = null;        return (new_node);    }     /* Function to print nodes in a given doubly linked list */    static void printList() {        while (head != null) {            Console.Write(head.data + " ");            head = head.right;        }    }     /* Driver code */    public static void Main(String[] args) {         // Let us create the tree shown in above diagram        node root = newNode(10);        root.left = newNode(12);        root.right = newNode(15);        root.left.left = newNode(25);        root.left.right = newNode(30);        root.right.left = newNode(36);        head = null;        tail = null;         // Convert to DLL        bintree2list(root);         // Print the converted list        printList();     }} // This code is contributed by gauravrajput1
<script>// A javascript program for in-place// conversion of Binary Tree to DLL     /*     * A binary tree node has data, and left and right pointers     */     class Node {        constructor() {            this.data = 0;            this.left = null;            this.right = null;        }    }    /*     * This is the core function to convert Tree to list.     */     var head, tail;     function bintree2listUtil(root) {        if (root == null)            return;         var left = root.left;        var right = root.right;         bintree2listUtil(root.left);         if (head == null) {            head = root;        }         root.left = tail;         if (tail != null) {            (tail).right = root;        }         tail = root;         bintree2listUtil(root.right);    }     // The main function that first calls    // bintree2listUtil()    function bintree2list( root) {         // Base case        if (root == null)            head = root;         bintree2listUtil(root);     }     /*     * Helper function that allocates a new node with the given data and null left     * and right pointers.     */     function newNode(data) {        var new_node = new Node();        new_node.data = data;        new_node.left = new_node.right = null;        return (new_node);    }     /* Function to print nodes in a given doubly linked list */    function printList() {        while (head != null) {            document.write(head.data + " ");            head = head.right;        }    }     /* Driver code */             // Let us create the tree shown in above diagram        var root = newNode(10);        root.left = newNode(12);        root.right = newNode(15);        root.left.left = newNode(25);        root.left.right = newNode(30);        root.right.left = newNode(36);        head = null;        tail = null;         // Convert to DLL        bintree2list(root);         // Print the converted list        printList(); // This code is contributed by gauravrajput1</script>
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This article is compiled by Ashish Mangla and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.You may also like to see Convert a given Binary Tree to Doubly Linked List | Set 2 for another simple and efficient solution. 
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