# 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

Recommended Practice

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 ` `#include ` `#include ` `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 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 sortedInorder` `            ``= ``new` `ArrayList();`   `        ``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 ` `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 ` `#include `   `/* 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

 ``

## PHP

 `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 ` `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).

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