Given a Binary Search Tree (BST) and a range, count number of nodes that lie in the given range.

Examples:

Input: 10 / \ 5 50 / / \ 1 40 100 Range: [5, 45] Output: 3 There are three nodes in range, 5, 10 and 40

Source: Google Question

The idea is to traverse the given binary search tree starting from root. For every node being visited, check if this node lies in range, if yes, then add 1 to result and recur for both of its children. If current node is smaller than low value of range, then recur for right child, else recur for left child.

Below is the implementation of above idea.

## C++

// C++ program to count BST nodes withing a given range #include<bits/stdc++.h> using namespace std; // A BST node struct node { int data; struct node* left, *right; }; // Utility function to create new node node *newNode(int data) { node *temp = new node; temp->data = data; temp->left = temp->right = NULL; return (temp); } // Returns count of nodes in BST in range [low, high] int getCount(node *root, int low, int high) { // Base case if (!root) return 0; // Special Optional case for improving efficiency if (root->data == high && root->data == low) return 1; // If current node is in range, then include it in count and // recur for left and right children of it if (root->data <= high && root->data >= low) return 1 + getCount(root->left, low, high) + getCount(root->right, low, high); // If current node is smaller than low, then recur for right // child else if (root->data < low) return getCount(root->right, low, high); // Else recur for left child else return getCount(root->left, low, high); } // Driver program int main() { // Let us construct the BST shown in the above figure node *root = newNode(10); root->left = newNode(5); root->right = newNode(50); root->left->left = newNode(1); root->right->left = newNode(40); root->right->right = newNode(100); /* Let us constructed BST shown in above example 10 / \ 5 50 / / \ 1 40 100 */ int l = 5; int h = 45; cout << "Count of nodes between [" << l << ", " << h << "] is " << getCount(root, l, h); return 0; }

## Java

// Java code to count BST nodes that // lie in a given range class BinarySearchTree { /* Class containing left and right child of current node and key value*/ static class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; } } // Root of BST Node root; // Constructor BinarySearchTree() { root = null; } // Returns count of nodes in BST in // range [low, high] int getCount(Node node, int low, int high) { // Base Case if(node == null) return 0; // If current node is in range, then // include it in count and recur for // left and right children of it if(node.data >= low && node.data <= high) return 1 + this.getCount(node.left, low, high)+ this.getCount(node.right, low, high); // If current node is smaller than low, // then recur for right child else if(node.data < low) return this.getCount(node.right, low, high); // Else recur for left child else return this.getCount(node.left, low, high); } // Driver function public static void main(String[] args) { BinarySearchTree tree = new BinarySearchTree(); tree.root = new Node(10); tree.root.left = new Node(5); tree.root.right = new Node(50); tree.root.left.left = new Node(1); tree.root.right.left = new Node(40); tree.root.right.right = new Node(100); /* Let us constructed BST shown in above example 10 / \ 5 50 / / \ 1 40 100 */ int l=5; int h=45; System.out.println("Count of nodes between [" + l + ", " + h+ "] is " + tree.getCount(tree.root, l, h)); } } // This code is contributed by Kamal Rawal

Output:

Count of nodes between [5, 45] is 3

Time complexity of the above program is O(h + k) where h is height of BST and k is number of nodes in given range.

This article is contributed by Gaurav Ahirwar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.