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 nodewith the given data and NULL left and rightpointers. */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 inthe correct place in the tree. Returns the newroot pointer which the caller should then use(the standard trick to avoid using referenceparameters). */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 inthe 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 childand 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 Listclass 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 childclass 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 childclass 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,RIGHTfunction 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 codelet 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 childand a pointer to right child */struct node { int data; struct node* left; struct node* right;};/* Helper function that allocates a new nodewith the given data and NULL left and rightpointers. */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 inthe correct place in the tree. Returns the newroot pointer which the caller should then use(the standard trick to avoid using referenceparameters). */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 thattree. Note that the entire tree does not needto 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 nodewith the given data and NULL left and rightpointers. */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 inthe correct place in the tree. Returns the newroot pointer which the caller should then use(the standard trick to avoid using referenceparameters). */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 thattree. Note that the entire tree does not needto 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 nodeclass 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 nodeclass 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 codeif __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// Treeusing System;// C# program to find minimum value node in Binary Search// Tree// A binary tree nodepublic 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 treeclass 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 $this->node; } }// Find the node with minimum value // in a Binary Search Treefunction get_minimum_value($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 BSTclass 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 BSTint 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 codeint 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 approachpublic 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 BSTclass Node: def __init__(self, data): self.data = data self.left = None self.right = None# Function to find the minimum element in a BSTdef 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 BSTroot = 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 BSTmin_val = find_minimum(root)# Print the minimum elementprint("Minimum element in the BST is:", min_val) |
C#
using System;// Node class for BSTpublic 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 BSTclass Node { constructor(data) { this.data = data; this.left = null; this.right = null; }}// Function to find the minimum element in a BSTfunction 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 BSTconst 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 BSTconst minVal = findMinimum(root);// Print the minimum elementconsole.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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