Smallest number in BST which is greater than or equal to N

Given a Binary Search Tree and a number N, the task is to find the smallest number in the binary search tree that is greater than or equal to N. Print the value of the element if it exists otherwise print -1.


Input: N = 20
Output: 21
Explanation: 21 is the smallest element greater than 20.

Input: N = 18
Output: 19
Explanation: 19 is the smallest element greater than 18.

The idea is to follow the recursive approach for solving the problem i.e. start searching for the element from the root.

  • If there is a leaf node having a value less than N, then element doesn’t exist and return -1.
  • Otherwise, if node’s value is greater than or equal to N and left child is NULL or less than N then return the node value.
  • Else if node’s value is less than N, then search for the element in the right subtree.
  • Else search for the element in the left subtree by calling the function recursively according to the left or right value.
// C++ program to find the smallest value 
// greater than or equal to N
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int data;
    Node *left, *right;

// To create new BST Node
Node* createNode(int item)
    Node* temp = new Node;
    temp->data = item;
    temp->left = temp->right = NULL;

    return temp;

// To add a new node in BST
Node* add(Node* node, int key)
    // if tree is empty return new node
    if (node == NULL)
        return createNode(key);

    // if key is less then or grater then
    // node value then recur down the tree
    if (key < node->data)
        node->left = add(node->left, key);
    else if (key > node->data)
        node->right = add(node->right, key);

    // return the (unchanged) node pointer
    return node;

// function to find min value less then N
int findMinforN(Node* root, int N)
    // If leaf node reached and is smaller than N
    if (root->left == NULL && root->right == NULL 
                                && root->data < N)
        return -1;

    // If node's value is greater than N and left value
    // is NULL or smaller then return the node value
    if ((root->data >= N && root->left == NULL) 
        || (root->data >= N && root->left->data < N))
        return root->data;

    // if node value is smaller than N search in the
    // right subtree
    if (root->data <= N)
        return findMinforN(root->right, N);

    // if node value is greater than N search in the
    // left subtree
        return findMinforN(root->left, N);

// Drivers code
int main()
    /*    19
        /    \
       7     21
     /   \
    3     11
         /   \
         9    14

    Node* root = NULL;
    root = add(root, 19);
    root = add(root, 7);
    root = add(root, 3);
    root = add(root, 11);
    root = add(root, 9);
    root = add(root, 13);
    root = add(root, 21);

    int N = 18;
    cout << findMinforN(root, N) << endl;

    return 0;


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