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Largest number less than or equal to N in BST (Iterative Approach)

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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. 
 

BST

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 have discussed recursive approach in below post. 
Largest number in BST which is less than or equal to N
Here an iterative approach is discussed. We try to find the predecessor of the target. Keep two pointers, one pointing to the current node and one for storing the answer. If the current node’s data > N, we move towards left. In other case, when current node’s data is less than N, the current node can be our answer (so far), and we move towards right. 

Implementation:

C++




// 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;
}
 
// Returns largest value smaller than or equal to
// key. If key is smaller than the smallest, it
// returns -1.
int findFloor(Node* root, int key)
{
    Node *curr = root, *ans = NULL;
    while (curr) {
        if (curr->key <= key) {
            ans = curr;
            curr = curr->right;
        }
        else
            curr = curr->left;
    }
    if (ans)
        return ans->key;
    return -1;
}
 
// Driver code
int main()
{
    int N = 25;
 
    Node* root = insert(root, 19);
    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", findFloor(root, N));
 
    return 0;
}


Java




// 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 = 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;
}
 
// Returns largest value smaller than or equal to
// key. If key is smaller than the smallest, it
// returns -1.
static int findFloor(Node root, int key)
{
    Node curr = root, ans = null;
    while (curr != null)
    {
        if (curr.key <= key)
        {
            ans = curr;
            curr = curr.right;
        }
        else
            curr = curr.left;
    }
    if (ans != null)
        return ans.key;
    return -1;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 25;
 
    Node root = new Node();
    insert(root, 19);
    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.printf("%d", findFloor(root, N));
}
}
 
/* This code is contributed by PrinciRaj1992 */


Python3




# Python3 code to find the largest value
# smaller than or equal to N
 
class newNode:
     
    def __init__(self, item):
         
        self.key = item
        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
 
# Returns largest value smaller than or
# equal to key. If key is smaller than
# the smallest, it returns -1.
def findFloor(root, key):
     
    curr = root
    ans = None
     
    while (curr):
        if (curr.key <= key):
            ans = curr
            curr = curr.right
        else:
            curr = curr.left
    if (ans):
        return ans.key
         
    return -1
 
# Driver code
if __name__ == '__main__':
     
    N = 25
    root = None
 
    root = insert(root, 19)
    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(findFloor(root, N))
 
# This code is contributed by bgangwar59


C#




// C# code to find the largest value smaller
// than or equal to N
using System;
using System.Collections.Generic;
     
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;
}
 
// Returns largest value smaller than or equal to
// key. If key is smaller than the smallest, it
// returns -1.
static int findFloor(Node root, int key)
{
    Node curr = root, ans = null;
    while (curr != null)
    {
        if (curr.key <= key)
        {
            ans = curr;
            curr = curr.right;
        }
        else
            curr = curr.left;
    }
    if (ans != null)
        return ans.key;
    return -1;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 25;
 
    Node root = new Node();
    insert(root, 19);
    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.Write("{0}", findFloor(root, N));
}
}
 
// This code is contributed by Rajput-Ji


Javascript




<script>
 
// Javascript code to find the largest
// value smaller than or equal to N
class Node
{
    constructor(item)
    {
        this.key = item;
        this.left = null;
        this.right = null;
    }
}
 
// To create new BST Node
function newNode(item)
{
    let temp = new Node(item);
    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;
}
 
// Returns largest value smaller than or
// equal to key. If key is smaller than
// the smallest, it returns -1.
function findFloor(root, key)
{
    let curr = root, ans = null;
    while (curr != null)
    {
        if (curr.key <= key)
        {
            ans = curr;
            curr = curr.right;
        }
        else
            curr = curr.left;
    }
    if (ans != null)
        return ans.key;
    return -1;
}
 
// Driver code
let N = 25;
 
let root = new Node(N);
insert(root, 19);
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(findFloor(root, N));
 
// This code is contributed by divyeshrabadiya07
 
</script>


Output

25


Last Updated : 22 Jul, 2022
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