Write a function to find the node with minimum value in a Binary Search Tree.
Example:
Input:
Output: 8
Input:
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++ 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;
} |
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 ));
}
} |
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]}" )
|
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]);
}
} |
// 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]);
|
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++ 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 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 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 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# 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 |
<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 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 ?> |
Minimum value in BST is 1
Time Complexity: O(h) where h is the height of Binary Search Tree
Auxiliary Space: O(1)