Find the node with minimum value in a Binary Search Tree

Write a function to find the node with minimum value in a Binary Search Tree.

Example:

Input: 

first example BST

Output: 8

Input: 

second example BST

Output: 10 

Brute Force Approach: To solve the problem follow the below idea:

The in-order traversal of a binary search tree always returns the value of nodes in sorted order. So the 1st value in the sorted vector will be the minimum value  which is the answer.

Below is the implementation of the above approach:

C++14

// C++ program to find minimum value node in binary search
// Tree.

#include <stdio.h>
#include <stdlib.h>
#include <vector>
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;
    }
}

/* Given a non-empty binary search tree,
inorder traversal for the tree is stored in
the vector sortedInorder.
Inorder is LEFT,ROOT,RIGHT*/
void inorder(struct node* node, vector<int>& sortedInorder)
{
    if (node == NULL)
        return;
    /* first recur on left child */
    inorder(node->left, sortedInorder);

    /* then insert the data of node */
    sortedInorder.push_back(node->data);

    /* now recur on right child */
    inorder(node->right, sortedInorder);
}

/* Driver code*/
int main()
{
    struct node* root = NULL;
    root = insert(root, 4);
    insert(root, 2);
    insert(root, 1);
    insert(root, 3);
    insert(root, 6);
    insert(root, 4);
    insert(root, 5);

    vector<int> sortedInorder;
    inorder(
        root,
        sortedInorder); // calling the recursive function
    // values of all nodes will appear in sorted order in
    // the vector sortedInorder
    // Function call
    printf("\n Minimum value in BST is %d",
           sortedInorder[0]);
    getchar();
    return 0;
}

Java

import java.util.*;

/* A binary tree node has data, pointer to left child
and a pointer to right child */
class Node {
    int data;
    Node left, right;

    /* Helper function that allocates a new node
    with the given data and NULL left and right
    pointers. */
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

class Main {

    /* 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). */
    public static 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 {

            /* 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;
        }
    }

    /* Given a non-empty binary search tree,
    inorder traversal for the tree is stored in
    the vector sortedInorder.
    Inorder is LEFT,ROOT,RIGHT*/
    public static void inorder(Node node,
                               List<Integer> sortedInorder)
    {
        if (node == null) {
            return;
        }
        /* first recur on left child */
        inorder(node.left, sortedInorder);

        /* then insert the data of node */
        sortedInorder.add(node.data);

        /* now recur on right child */
        inorder(node.right, sortedInorder);
    }

    public static void main(String[] args)
    {
        Node root = null;
        root = insert(root, 4);
        insert(root, 2);
        insert(root, 1);
        insert(root, 3);
        insert(root, 6);
        insert(root, 4);
        insert(root, 5);

        List<Integer> sortedInorder
            = new ArrayList<Integer>();

        inorder(root, sortedInorder); // calling the
                                      // recursive function
        // values of all nodes will appear in sorted order
        // in the vector sortedInorder
        // Function call
        System.out.printf("\n Minimum value in BST is %d",
                          sortedInorder.get(0));
    }
}

Python3

from typing import List


class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# 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).


def insert(node: Node, data: int) -> Node:

    # If the tree is empty, return a new, single node
    if not node:
        return Node(data)

    # 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

# Given a non-empty binary search tree, inorder traversal for
# the tree is stored in the list sorted_inorder. Inorder is LEFT,ROOT,RIGHT.


def inorder(node: Node, sorted_inorder: List[int]) -> None:

    if not node:
        return

    # First recur on left child
    inorder(node.left, sorted_inorder)

    # Then insert the data of node
    sorted_inorder.append(node.data)

    # Now recur on right child
    inorder(node.right, sorted_inorder)


if __name__ == '__main__':
    root = None
    root = insert(root, 4)
    insert(root, 2)
    insert(root, 1)
    insert(root, 3)
    insert(root, 6)
    insert(root, 4)
    insert(root, 5)

    sorted_inorder = []
    inorder(root, sorted_inorder)  # calling the recursive function

    # Values of all nodes will appear in sorted order in the list sorted_inorder
    print(f"Minimum value in BST is {sorted_inorder[0]}")

C#

using System;
using System.Collections.Generic;

// A binary tree node has data, pointer to left child,
// and a pointer to right child
class Node {
    public int data;
    public Node left, right;

    // Helper function that allocates a new node
    // with the given data and NULL left and right
    // pointers.
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

class MainClass {
    // 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).
    public static 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 {
            // 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;
        }
    }

    // Given a non-empty binary search tree,
    // inorder traversal for the tree is stored in
    // the List sortedInorder.
    // Inorder is LEFT,ROOT,RIGHT
    public static void inorder(Node node,
                               List<int> sortedInorder)
    {
        if (node == null) {
            return;
        }

        // first recur on left child
        inorder(node.left, sortedInorder);

        // then insert the data of node
        sortedInorder.Add(node.data);

        // now recur on right child
        inorder(node.right, sortedInorder);
    }

    public static void Main(string[] args)
    {
        Node root = null;
        root = insert(root, 4);
        insert(root, 2);
        insert(root, 1);
        insert(root, 3);
        insert(root, 6);
        insert(root, 4);
        insert(root, 5);

        List<int> sortedInorder = new List<int>();

        inorder(root, sortedInorder); // calling the
                                      // recursive function

        // values of all nodes will appear in sorted order
        // in the list sortedInorder
        // Function call
        Console.WriteLine("\n Minimum value in BST is {0}",
                          sortedInorder[0]);
    }
}

Javascript

// A binary tree node has data, pointer to left child
// and a pointer to right child
class Node {
    constructor(data) {
        this.data = data;
        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) {
    return new Node(data);
}

// 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;
    }
}

// Given a non-empty binary search tree,
// inorder traversal for the tree is stored in
// the vector sortedInorder.
// Inorder is LEFT,ROOT,RIGHT
function inorder(node, sortedInorder) {
    if (node == null) return;
    // first recur on left child
    inorder(node.left, sortedInorder);

    // then insert the data of node
    sortedInorder.push(node.data);

    // now recur on right child
    inorder(node.right, sortedInorder);
}

// Driver code
let root = null;
root = insert(root, 4);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 6);
insert(root, 4);
insert(root, 5);

let sortedInorder = [];
inorder(root, sortedInorder);
console.log("Minimum value in BST is " + sortedInorder[0]);

Output

 Minimum value in BST is 1

Time Complexity: O(n) where n is the number of nodes.
Auxiliary Space: O(n) (for recursive stack space + vector used additionally)

Efficient Approach: To solve the problem follow the below idea:

This is quite simple. Just traverse the node from root to left recursively until left is NULL. The node whose left is NULL is the node with minimum value

Below is the implementation of the above approach:

C++

// C++ program to find minimum value node in binary search
// Tree.
#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;
    }
}

/* 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. */
int minValue(struct node* node)
{
    struct node* current = node;

    /* loop down to find the leftmost leaf */
    while (current->left != NULL) {
        current = current->left;
    }
    return (current->data);
}

/* Driver Code*/
int main()
{
    struct node* root = NULL;
    root = insert(root, 4);
    insert(root, 2);
    insert(root, 1);
    insert(root, 3);
    insert(root, 6);
    insert(root, 5);

      // Function call
    cout << "\n Minimum value in BST is " << minValue(root);
    getchar();
    return 0;
}

// This code is contributed by Mukul Singh.

C

// C program to find minimum value node in binary search
// Tree.

#include <stdio.h>
#include <stdlib.h>

/* 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;
    }
}

/* 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. */
int minValue(struct node* node)
{
    struct node* current = node;

    /* loop down to find the leftmost leaf */
    while (current->left != NULL) {
        current = current->left;
    }
    return (current->data);
}

/* Driver code*/
int main()
{
    struct node* root = NULL;
    root = insert(root, 4);
    insert(root, 2);
    insert(root, 1);
    insert(root, 3);
    insert(root, 6);
    insert(root, 5);

      // Function call
    printf("\n Minimum value in BST is %d", minValue(root));
    getchar();
    return 0;
}

Java

// Java program to find minimum value node in Binary Search
// Tree

// A binary tree node
class Node {

    int data;
    Node left, right;

    Node(int d)
    {
        data = d;
        left = right = 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 {

            /* 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;
        }
    }

    /* 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. */
    int minvalue(Node node)
    {
        Node current = node;

        /* loop down to find the leftmost leaf */
        while (current.left != null) {
            current = current.left;
        }
        return (current.data);
    }

    // Driver code
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        Node root = null;
        root = tree.insert(root, 4);
        tree.insert(root, 2);
        tree.insert(root, 1);
        tree.insert(root, 3);
        tree.insert(root, 6);
        tree.insert(root, 5);

          // Function call
        System.out.println("Minimum value of BST is "
                           + tree.minvalue(root));
    }
}

// This code is contributed by Mayank Jaiswal

Python3

# Python3 program to find the node with minimum value in bst

# A binary tree node


class Node:

    # Constructor to create a new node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None


""" 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). """


def insert(node, data):

    # 1. If the tree is empty, return a new,
    # single node
    if node is None:
        return (Node(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


""" 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. """


def minValue(node):
    current = node

    # loop down to find the leftmost leaf
    while(current.left is not None):
        current = current.left

    return current.data


# Driver code
if __name__ == '__main__':
  root = None
  root = insert(root, 4)
  insert(root, 2)
  insert(root, 1)
  insert(root, 3)
  insert(root, 6)
  insert(root, 5)

  # Function call
  print("\nMinimum value in BST is %d" % (minValue(root)))

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# program to find minimum value node in Binary Search
// Tree

using System;

// C# program to find minimum value node in Binary Search
// Tree

// A binary tree node
public class Node {

    public int data;
    public Node left, right;

    public Node(int d)
    {
        data = d;
        left = right = null;
    }
}

public class BinaryTree {

    public 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). */
    public virtual 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 {

            /* 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;
        }
    }

    /* 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. */
    public virtual int minvalue(Node node)
    {
        Node current = node;

        /* loop down to find the leftmost leaf */
        while (current.left != null) {
            current = current.left;
        }
        return (current.data);
    }

    // Driver code
    public static void Main(string[] args)
    {
        BinaryTree tree = new BinaryTree();
        Node root = null;
        root = tree.insert(root, 4);
        tree.insert(root, 2);
        tree.insert(root, 1);
        tree.insert(root, 3);
        tree.insert(root, 6);
        tree.insert(root, 5);

          // Function call
        Console.WriteLine("Minimum value of BST is "
                          + tree.minvalue(root));
    }
}

// This code is contributed by Shrikant13

Javascript

<script>

    // JavaScript program to find minimum
    // value node in Binary Search Tree

    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    let 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). */
    function insert(node, data) {

        /* 1. If the tree is empty, return a new,
         single node */
        if (node == null) {
            return (new Node(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;
        }
    }

    /* 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) {
        if (node === null) return null;
        let current = node;

        /* loop down to find the leftmost leaf */
        while (current.left != null) {
            current = current.left;
        }
        return (current.data);
    }

    let root = null;
    root = insert(root, 4);
    insert(root, 2);
    insert(root, 1);
    insert(root, 3);
    insert(root, 6);
    insert(root, 5);

    document.write("Minimum value in BST is " + minvalue(root));

</script>

PHP

<?php
// PHP program to find the node with
// minimum value in bst

// create a binary tree
class node
{
    private $node, $left, $right;
    function __construct($node)
    {
        $this->node = $node;
        $left = $right = NULL;
    }

    // set the left node in tree
    function set_left($left)
    {
        $this->left = $left;
    }

    // set the right node in tree
    function set_right($right)
    {
        $this->right = $right;
    }

    // get left node
    function get_left()
    {
        return $this->left;
    }

    // get right node
    function get_right()
    {
        return $this->right;
    }

    // get value of current node
    function get_node()
    {
        return $this->node;
    }

}

// Find the node with minimum value
// in a Binary Search Tree
function get_minimum_value($node)
{
    /*travel till last left node to
      get the minimum value*/
    while ($node->get_left() != NULL)
    {
        $node = $node->get_left();
    }
    return $node->get_node();
}

// code to creating a tree
$node = new node(4);
$lnode = new node(2);
$lnode->set_left(new node(1));
$lnode->set_right(new node(3));
$rnode = new node(6);
$rnode->set_left(new node(5));
$node->set_left($lnode);
$node->set_right($rnode);

$minimum_value = get_minimum_value($node);
echo 'Minimum value of BST is '.
                 $minimum_value;

// This code is contributed
// by Deepika Pathak
?>

Output

 Minimum value in BST is 1

Time Complexity: O(n) where n is the number of nodes.
Auxiliary Space: O(1)

Another approach Modified binary search approach:

The basic idea behind this approach is to exploit the properties of a Binary Search Tree (BST). In a BST, the left subtree of a node contains all the nodes with values less than the node's value, and the right subtree contains all the nodes with values greater than the node's value.

• Follow the steps to implement the above idea:
• Start at the root node of the BST.
• If the left child of the current node is NULL, return the value of the current node. This is the minimum element in the BST.
• If the value of the left child is less than the value of the current node, move to the left subtree and repeat step 2.
• If the value of the left child is greater than or equal to the value of the current node, move to the right subtree and repeat step 2.
• Repeat steps 2-4 until you reach a leaf node.

Below is the implementation of the above approach:

C++

// CPP program to implement the modified binary search approach
#include <iostream>
using namespace std;

// Node class for BST
class Node {
public:
    int data;
    Node* left;
    Node* right;

    Node(int data) {
        this->data = data;
        left = right = nullptr;
    }
};

// Function to find the minimum element in a BST
int findMinimum(Node* root) {
    if (root == nullptr) {
        return -1;
    }
    while (root->left != nullptr) {    // Traverse to the leftmost node
        if (root->left->data < root->data) {
            root = root->left;
        } else {
            root = root->right;
        }
    }
    return root->data;
}

// Driver code
int main() {
    // Create a BST
    Node* root = new Node(4);
    root->left = new Node(2);
    root->right = new Node(6);
    root->left->left = new Node(1);
    root->left->right = new Node(3);
    root->right->left = new Node(5);
    root->right->right = new Node(7);

    // Find the minimum element in the BST
    int minVal = findMinimum(root);

    // Print the minimum element
    cout << "Minimum element in the BST is: " << minVal << endl;

    return 0;
}

// This code is contributed by Susobhan Akhuli

Java

// Java program to implement the modified binary search approach
public class Main {

    // Node class for BST
    static class Node {
        int data;
        Node left, right;

        Node(int data) {
            this.data = data;
            left = right = null;
        }
    }

    // Function to find the minimum element in a BST
    static int findMinimum(Node root) {
        if (root == null) {
            return -1;
        }
        while (root.left != null) {    // Traverse to the leftmost node
            if (root.left.data < root.data) {
                root = root.left;
            } else {
                root = root.right;
            }
        }
        return root.data;
    }

    // Driver code
    public static void main(String[] args) {
        // Create a BST
        Node root = new Node(4);
        root.left = new Node(2);
        root.right = new Node(6);
        root.left.left = new Node(1);
        root.left.right = new Node(3);
        root.right.left = new Node(5);
        root.right.right = new Node(7);

        // Find the minimum element in the BST
        int minVal = findMinimum(root);

        // Print the minimum element
        System.out.println("Minimum element in the BST is: " + minVal);
    }
}

Python3

# Node class for BST
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# Function to find the minimum element in a BST
def find_minimum(root):
    if root is None:
        return -1
    while root.left is not None:  # Traverse to the leftmost node
        if root.left.data < root.data:
            root = root.left
        else:
            root = root.right
    return root.data

# Driver code
# Create a BST
root = Node(4)
root.left = Node(2)
root.right = Node(6)
root.left.left = Node(1)
root.left.right = Node(3)
root.right.left = Node(5)
root.right.right = Node(7)

# Find the minimum element in the BST
min_val = find_minimum(root)

# Print the minimum element
print("Minimum element in the BST is:", min_val)

C#

using System;

// Node class for BST
public class Node
{
    public int data;
    public Node left;
    public Node right;

    public Node(int val)
    {
        data = val;
        left = null;
        right = null;
    }
}

public class GFG
{
    // Function to find the minimum element in a BST
    public static int FindMinimum(Node root)
    {
        if (root == null)
        {
            return -1;
        }

        while (root.left != null)
        {
            if (root.left.data < root.data)
            {
                root = root.left;
            }
            else
            {
                root = root.right;
            }
        }

        return root.data;
    }

    public static void Main(string[] args)
    {
        // Create a BST
        Node root = new Node(4);
        root.left = new Node(2);
        root.right = new Node(6);
        root.left.left = new Node(1);
        root.left.right = new Node(3);
        root.right.left = new Node(5);
        root.right.right = new Node(7);

        // Find the minimum element in the BST
        int minVal = FindMinimum(root);

        // Print the minimum element
        Console.WriteLine("Minimum element in the BST is: " + minVal);
    }
}

Javascript

// Node class for BST
class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

// Function to find the minimum element in a BST
function findMinimum(root) {
  if (root === null) {
    return -1;
  }
  while (root.left !== null) { // Traverse to the leftmost node
    if (root.left.data < root.data) {
      root = root.left;
    } else {
      root = root.right;
    }
  }
  return root.data;
}

// Driver code
// Create a BST
const root = new Node(4);
root.left = new Node(2);
root.right = new Node(6);
root.left.left = new Node(1);
root.left.right = new Node(3);
root.right.left = new Node(5);
root.right.right = new Node(7);

// Find the minimum element in the BST
const minVal = findMinimum(root);

// Print the minimum element
console.log("Minimum element in the BST is:", minVal);

Output

Minimum element in the BST is: 1

Time Complexity: O(log n) , This approach has a time complexity of O(log n), where n is the number of nodes in the BST.
space complexity: O(1).