Find the node with minimum value in a Binary Search Tree using recursion - GeeksforGeeks

23 Jun, 2021
Given a Binary Search Tree, the task is to find the node with minimum value.
Examples: 
Input:  
Output: 4   
Approach: Just traverse the node from root to left recursively until left is NULL. The node whose left is NULL is the node with the minimum value.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; /* A binary tree node has data, pointer to left child   and a pointer to right child */struct node {    int data;    struct node* left;    struct node* right;}; /* 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;     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 {         /* 2. Otherwise, recur down the tree */        if (data <= node->data)            node->left = insert(node->left, data);        else            node->right = insert(node->right, data);         /* return the (unchanged) node pointer */        return node;    }} // Function to return the minimum node// in the given binary search treeint minValue(struct node* node){    if (node->left == NULL)        return node->data;    return minValue(node->left);} // Driver codeint main(){     // Create the BST    struct node* root = NULL;    root = insert(root, 4);    insert(root, 2);    insert(root, 1);    insert(root, 3);    insert(root, 6);    insert(root, 5);     cout << minValue(root);     return 0;}
// Java Implementation of the above approach class GFG{     /* A binary tree node has data, pointer to left childand a pointer to right child */static class Node{    int data;    Node left;    Node right;}; /* Helper function that allocates a new nodewith the given data and null left and rightpointers. */static Node newNode(int data){    Node node = new Node();    node.data = data;    node.left = null;    node.right = null;     return (node);} /* Give a binary search tree and a number,inserts a new node with the given number inthe correct place in the tree. Returns the newroot pointer which the caller should then use(the standard trick to avoid using referenceparameters). */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    {         /* 2. Otherwise, recur down the tree */        if (data <= node.data)            node.left = insert(node.left, data);        else            node.right = insert(node.right, data);         /* return the (unchanged) node pointer */        return node;    }} // Function to return the minimum node// in the given binary search treestatic int minValue(Node node){    if (node.left == null)        return node.data;    return minValue(node.left);} // Driver codepublic static void main(String args[]){    // Create the BST    Node root = null;    root = insert(root, 4);    insert(root, 2);    insert(root, 1);    insert(root, 3);    insert(root, 6);    insert(root, 5);     System.out.println(minValue(root)); }} // This code has been contributed by 29AjayKumar
# Python3 Implementation of# the above approachclass Node:    def __init__(self, data):        self.data = data        self.left = None        self.right = None # Helper function that allocates # a new node with the given data# and null left and right pointers.def insert(node, data):    if node is None :        return Node(data)    else:        if data <= node.data:            node.left = insert(node.left, data)        else:            node.right = insert(node.right, data)         return node # Function to return the minimum node# in the given binary search treedef minValue(node):    if node.left == None:        return node.data    return minValue(node.left) # Driver codeif __name__ == "__main__" :         # Create the BST    root = None    root = insert(root, 4)    insert(root, 2)    insert(root, 1)    insert(root, 3)    insert(root, 6)    insert(root, 5)     print(minValue(root)) # This code is contributed by vinayak
// C# Implementation of the above approachusing System;     class GFG{     /* A binary tree node has data, pointer to left childand a pointer to right child */public class Node{    public int data;    public Node left;    public Node right;}; /* Helper function that allocates a new nodewith the given data and null left and rightpointers. */static Node newNode(int data){    Node node = new Node();    node.data = data;    node.left = null;    node.right = null;     return (node);} /* Give a binary search tree and a number,inserts a new node with the given number inthe correct place in the tree. Returns the newroot pointer which the caller should then use(the standard trick to avoid using referenceparameters). */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    {         /* 2. Otherwise, recur down the tree */        if (data <= node.data)            node.left = insert(node.left, data);        else            node.right = insert(node.right, data);         /* return the (unchanged) node pointer */        return node;    }} // Function to return the minimum node// in the given binary search treestatic int minValue(Node node){    if (node.left == null)        return node.data;    return minValue(node.left);} // Driver codepublic static void Main(String []args){    // Create the BST    Node root = null;    root = insert(root, 4);    insert(root, 2);    insert(root, 1);    insert(root, 3);    insert(root, 6);    insert(root, 5);     Console.WriteLine(minValue(root)); }} // This code contributed by Rajput-Ji
<script> // Javascript implementation of the above approach     // A binary tree node has data, pointer to// left child and a pointer to right childclass Node{    constructor()    {        this.data = 0;        this.left = null;        this.right = null;    }}; // 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;     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).function insert(node, data){         /* 1. If the tree is empty, return a new,    single node */    if (node == null)        return (newNode(data));    else    {                 /* 2. Otherwise, recur down the tree */        if (data <= node.data)            node.left = insert(node.left, data);        else            node.right = insert(node.right, data);         /* return the (unchanged) node pointer */        return node;    }} // Function to return the minimum node// in the given binary search treefunction minValue(node){    if (node.left == null)        return node.data;             return minValue(node.left);} // Driver code // Create the BSTvar root = null;root = insert(root, 4);insert(root, 2);insert(root, 1);insert(root, 3);insert(root, 6);insert(root, 5); document.write(minValue(root)); // This code is contributed by noob2000 </script>
1
 
Time Complexity: O(n), worst case happens for left skewed trees. 
29AjayKumar
Rajput-Ji
itsvinayak
noob2000
Traversal
Binary Search Tree
Tree
Traversal
Binary Search Tree
Tree
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