We have a binary search tree and a number N. Our task is to find the greatest number in the binary search tree that is less than or equal to N. Print the value of the element if it exists otherwise print -1.
Examples:
For the above given binary search tree-
Input : N = 24 Output :result = 21 (searching for 24 will be like-5->12->21) Input : N = 4 Output : result = 3 (searching for 4 will be like-5->2->3)
We follow recursive approach for solving this problem. We start searching for element from root node. If we reach a leaf and its value is greater than N, element does not exist so return -1. Else if node’s value is less than or equal to N and right value is NULL or greater than N, then return the node value as it will be the answer.
Otherwise if node’s value is greater than N, then search for the element in the left subtree else search for the element in the right subtree by calling the same function by passing the left or right values accordingly.
Implementation:
// C++ code to find the largest value smaller // than or equal to N #include <bits/stdc++.h> using namespace std;
struct Node {
int key;
Node *left, *right;
}; // To create new BST Node Node* newNode( int item)
{ Node* temp = new Node;
temp->key = item;
temp->left = temp->right = NULL;
return temp;
} // To insert a new node in BST Node* insert(Node* node, int key)
{ // if tree is empty return new node
if (node == NULL)
return newNode(key);
// if key is less than or greater than
// node value then recur down the tree
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
// return the (unchanged) node pointer
return node;
} // function to find max value less than N int findMaxforN(Node* root, int N)
{ // Base cases
if (root == NULL)
return -1;
if (root->key == N)
return N;
// If root's value is smaller, try in right
// subtree
else if (root->key < N) {
int k = findMaxforN(root->right, N);
if (k == -1)
return root->key;
else
return k;
}
// If root's key is greater, return value
// from left subtree.
else if (root->key > N)
return findMaxforN(root->left, N);
} // Driver code int main()
{ int N = 4;
// creating following BST
/*
5
/ \
2 12
/ \ / \
1 3 9 21
/ \
19 25 */
Node* root = insert(root, 25);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 12);
insert(root, 9);
insert(root, 21);
insert(root, 19);
insert(root, 25);
printf ( "%d" , findMaxforN(root, N));
return 0;
} |
// Java code to find the largest value smaller // than or equal to N class GfG {
static class Node {
int key;
Node left, right;
} // To create new BST Node static Node newNode( int item)
{ Node temp = new Node();
temp.key = item;
temp.left = null ;
temp.right = null ;
return temp;
} // To insert a new node in BST static Node insert(Node node, int key)
{ // if tree is empty return new node
if (node == null )
return newNode(key);
// if key is less than or greater than
// node value then recur down the tree
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
// return the (unchanged) node pointer
return node;
} // function to find max value less than N static int findMaxforN(Node root, int N)
{ // Base cases
if (root == null )
return - 1 ;
if (root.key == N)
return N;
// If root's value is smaller, try in right
// subtree
else if (root.key < N) {
int k = findMaxforN(root.right, N);
if (k == - 1 )
return root.key;
else
return k;
}
// If root's key is greater, return value
// from left subtree.
else if (root.key > N)
return findMaxforN(root.left, N);
return - 1 ;
} // Driver code public static void main(String[] args)
{ int N = 4 ;
// creating following BST
/*
5
/ \
2 12
/ \ / \
1 3 9 21
/ \
19 25 */
Node root = null ;
root = insert(root, 25 );
insert(root, 2 );
insert(root, 1 );
insert(root, 3 );
insert(root, 12 );
insert(root, 9 );
insert(root, 21 );
insert(root, 19 );
insert(root, 25 );
System.out.println(findMaxforN(root, N));
} } |
# Python3 code to find the largest # value smaller than or equal to N class newNode:
# Constructor to create a new node
def __init__( self , data):
self .key = data
self .left = None
self .right = None
# To insert a new node in BST def insert(node, key):
# if tree is empty return new node
if node = = None :
return newNode(key)
# if key is less than or greater than
# node value then recur down the tree
if key < node.key:
node.left = insert(node.left, key)
elif key > node.key:
node.right = insert(node.right, key)
# return the (unchanged) node pointer
return node
# function to find max value less than N def findMaxforN(root, N):
# Base cases
if root = = None :
return - 1
if root.key = = N:
return N
# If root's value is smaller, try in
# right subtree
elif root.key < N:
k = findMaxforN(root.right, N)
if k = = - 1 :
return root.key
else :
return k
# If root's key is greater, return
# value from left subtree.
elif root.key > N:
return findMaxforN(root.left, N)
# Driver code if __name__ = = '__main__' :
N = 4
# creating following BST
#
# 5
# / \
# 2 12
# / \ / \
# 1 3 9 21
# / \
# 19 25
root = None
root = insert(root, 25 )
insert(root, 2 )
insert(root, 1 )
insert(root, 3 )
insert(root, 12 )
insert(root, 9 )
insert(root, 21 )
insert(root, 19 )
insert(root, 25 )
print (findMaxforN(root, N))
# This code is contributed by PranchalK |
// C# code to find the largest value // smaller than or equal to N using System;
class GFG
{ class Node
{
public int key;
public Node left, right;
}
// To create new BST Node
static Node newNode( int item)
{
Node temp = new Node();
temp.key = item;
temp.left = null ;
temp.right = null ;
return temp;
}
// To insert a new node in BST
static Node insert(Node node, int key)
{
// if tree is empty return new node
if (node == null )
return newNode(key);
// if key is less than or greater than
// node value then recur down the tree
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
// return the (unchanged) node pointer
return node;
}
// function to find max value less than N
static int findMaxforN(Node root, int N)
{
// Base cases
if (root == null )
return -1;
if (root.key == N)
return N;
// If root's value is smaller,
// try in right subtree
else if (root.key < N)
{
int k = findMaxforN(root.right, N);
if (k == -1)
return root.key;
else
return k;
}
// If root's key is greater, return
// value from left subtree.
else if (root.key > N)
return findMaxforN(root.left, N);
return -1;
}
// Driver code
public static void Main(String[] args)
{
int N = 4;
// creating following BST
/*
5
/ \
2 12
/ \ / \
1 3 9 21
/ \
19 25 */
Node root = null ;
root = insert(root, 25);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 12);
insert(root, 9);
insert(root, 21);
insert(root, 19);
insert(root, 25);
Console.WriteLine(findMaxforN(root, N));
}
} // This code is contributed 29AjayKumar |
<script> // javascript code to find the largest value smaller // than or equal to N class Node
{
constructor(){
this .key = 0;
this .left = null , this .right = null ;
}
}
// To create new BST Node
function newNode(item)
{
var temp = new Node();
temp.key = item;
temp.left = null ;
temp.right = null ;
return temp;
}
// To insert a new node in BST
function insert(node , key) {
// if tree is empty return new node
if (node == null )
return newNode(key);
// if key is less than or greater than
// node value then recur down the tree
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
// return the (unchanged) node pointer
return node;
}
// function to find max value less than N
function findMaxforN(root , N)
{
// Base cases
if (root == null )
return -1;
if (root.key == N)
return N;
// If root's value is smaller, try in right
// subtree
else if (root.key < N)
{
var k = findMaxforN(root.right, N);
if (k == -1)
return root.key;
else
return k;
}
// If root's key is greater, return value
// from left subtree.
else if (root.key > N)
return findMaxforN(root.left, N);
return -1;
}
// Driver code
var N = 4;
// creating following BST
/*
* 5
/ \
2 12
/ \ / \
1 3 9 21
/ \
19 25
*/
var root = null ;
root = insert(root, 25);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 12);
insert(root, 9);
insert(root, 21);
insert(root, 19);
insert(root, 25);
document.write(findMaxforN(root, N));
// This code is contributed by Rajput-Ji </script> |
3
Time Complexity: O(h), where h is height of BST.
Auxiliary Space: O(h), The extra space is used in recursion call stack.
Iterative Solution: Below is an iterative solution and it does not require extra space for recursion call stack.
Implementation:
// C++ code to find the largest value smaller // than or equal to N #include <bits/stdc++.h> using namespace std;
struct Node {
int key;
Node *left, *right;
}; // To create new BST Node Node* newNode( int item)
{ Node* temp = new Node;
temp->key = item;
temp->left = temp->right = NULL;
return temp;
} // To insert a new node in BST Node* insert(Node* node, int key)
{ // if tree is empty return new node
if (node == NULL)
return newNode(key);
// if key is less than or greater than
// node value then recur down the tree
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
// return the (unchanged) node pointer
return node;
} // function to find max value less than N void findMaxforN(Node* root, int N)
{ // Start from root and keep looking for larger
while (root != NULL && root->right != NULL) {
// If root is smaller go to right side
if (N > root->key && N >= root->right->key)
root = root->right;
// If root is greater go to left side
else if (N < root->key)
root = root->left;
else
break ;
}
if (root == NULL || root->key > N)
cout << -1;
else
cout << root->key;
} // Driver code int main()
{ int N = 50;
Node* root = insert(root, 5);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 12);
insert(root, 9);
insert(root, 21);
insert(root, 19);
insert(root, 25);
findMaxforN(root, N);
return 0;
} |
// Java code to find the largest value smaller // than or equal to N import java.util.*;
class GFG{
static class Node {
int key;
Node left, right;
};
// To create new BST Node
static Node newNode( int item)
{
Node temp = new Node();
temp.key = item;
temp.left = temp.right = null ;
return temp;
}
// To insert a new node in BST
static Node insert(Node node, int key)
{
// if tree is empty return new node
if (node == null )
return newNode(key);
// if key is less than or greater than
// node value then recur down the tree
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
// return the (unchanged) node pointer
return node;
}
// function to find max value less than N
static void findMaxforN(Node root, int N)
{
// Start from root and keep looking for larger
while (root != null && root.right != null ) {
// If root is smaller go to right side
if (N > root.key && N >= root.right.key)
root = root.right;
// If root is greater go to left side
else if (N < root.key)
root = root.left;
else
break ;
}
if (root == null || root.key > N)
System.out.print(- 1 );
else
System.out.print(root.key);
}
// Driver code
public static void main(String[] args)
{
int N = 50 ;
Node root = insert( null , 5 );
insert(root, 2 );
insert(root, 1 );
insert(root, 3 );
insert(root, 12 );
insert(root, 9 );
insert(root, 21 );
insert(root, 19 );
insert(root, 25 );
findMaxforN(root, N);
}
} // This code is contributed by Rajput-Ji |
# Python3 code to find the largest value # smaller than or equal to N class newNode:
# To create new BST Node
def __init__( self , data):
self .key = data
self .left = None
self .right = None
# To insert a new node in BST def insert(node, key):
# If tree is empty return new node
if (node = = None ):
return newNode(key)
# If key is less than or greater than
# node value then recur down the tree
if (key < node.key):
node.left = insert(node.left, key)
elif (key > node.key):
node.right = insert(node.right, key)
# Return the (unchanged) node pointer
return node
# Function to find max value less than N def findMaxforN(root, N):
# Start from root and keep looking for larger
while (root ! = None and root.right ! = None ):
# If root is smaller go to right side
if (N > root.key and N > = root.right.key):
root = root.right
# If root is greater go to left side
elif (N < root.key):
root = root.left
else :
break
if (root = = None or root.key > N):
print ( - 1 )
else :
print (root.key)
# Driver code if __name__ = = '__main__' :
N = 50
root = None
root = insert(root, 5 )
insert(root, 2 )
insert(root, 1 )
insert(root, 3 )
insert(root, 12 )
insert(root, 9 )
insert(root, 21 )
insert(root, 19 )
insert(root, 25 )
findMaxforN(root, N)
# This code is contributed by bgangwar59 |
// C# code to find the largest value smaller // than or equal to N using System;
public class GFG {
public class Node {
public int key;
public Node left, right;
};
// To create new BST Node
static Node newNode( int item) {
Node temp = new Node();
temp.key = item;
temp.left = temp.right = null ;
return temp;
}
// To insert a new node in BST
static Node insert(Node node, int key) {
// if tree is empty return new node
if (node == null )
return newNode(key);
// if key is less than or greater than
// node value then recur down the tree
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
// return the (unchanged) node pointer
return node;
}
// function to find max value less than N
static void findMaxforN(Node root, int N) {
// Start from root and keep looking for larger
while (root != null && root.right != null ) {
// If root is smaller go to right side
if (N > root.key && N >= root.right.key)
root = root.right;
// If root is greater go to left side
else if (N < root.key)
root = root.left;
else
break ;
}
if (root == null || root.key > N)
Console.Write(-1);
else
Console.Write(root.key);
}
// Driver code
public static void Main(String[] args) {
int N = 50;
Node root = insert( null , 5);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 12);
insert(root, 9);
insert(root, 21);
insert(root, 19);
insert(root, 25);
findMaxforN(root, N);
}
} // This code is contributed by Rajput-Ji |
<script> // JavaScript code to find the largest value smaller // than or equal to N class Node { constructor(data)
{
this .key=data;
this .left= this .right= null ;
}
} function insert(node, key)
{ // If tree is empty return new node
if (node == null )
return new Node(key)
// If key is less than or greater than
// node value then recur down the tree
if (key < node.key)
node.left = insert(node.left, key)
else if (key > node.key)
node.right = insert(node.right, key)
// Return the (unchanged) node pointer
return node
} function findMaxforN(root, N)
{ // Start from root and keep looking for larger
while (root != null && root.right != null )
{
// If root is smaller go to right side
if (N > root.key && N >= root.right.key)
root = root.right
// If root is greater go to left side
else if (N < root.key)
root = root.left
else
break
}
if (root == null || root.key > N)
document.write(-1)
else
document.write(root.key)
} let N = 50; let root = null ;
root=insert(root, 2) root=insert(root, 1) root=insert(root, 3) root=insert(root, 12) root=insert(root, 9) root=insert(root, 21) root=insert(root, 19) root=insert(root, 25) findMaxforN(root, N) // This code is contributed by rag2127 </script> |
25
Time Complexity: O(h), where h is height of BST.
Auxiliary Space: O(1), As constant extra space is used.