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

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` |

## Python3

`# Python3 program to count BST nodes` `# withing a given range` `# Utility function to create new node` `class` `newNode:` ` ` `# Constructor to create a new node` ` ` `def` `__init__(` `self` `, data):` ` ` `self` `.data ` `=` `data` ` ` `self` `.left ` `=` `None` ` ` `self` `.right ` `=` `None` `# Returns count of nodes in BST in` `# range [low, high]` `def` `getCount(root, low, high):` ` ` ` ` `# Base case` ` ` `if` `root ` `=` `=` `None` `:` ` ` `return` `0` ` ` ` ` `# Special Optional case for improving` ` ` `# efficiency` ` ` `if` `root.data ` `=` `=` `high ` `and` `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 ` `and` `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` ` ` `elif` `root.data < low:` ` ` `return` `getCount(root.right, low, high)` ` ` `# Else recur for left child` ` ` `else` `:` ` ` `return` `getCount(root.left, low, high)` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Let us construct the BST shown in` ` ` `# the above figure` ` ` `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` ` ` `l ` `=` `5` ` ` `h ` `=` `45` ` ` `print` `(` `"Count of nodes between ["` `, l, ` `", "` `, h,` `"] is "` `,` ` ` `getCount(root, l, h))` `# This code is contributed by PranchalK` |

## C#

`using` `System;` `// C# code to count BST nodes that` `// lie in a given range` `public` `class` `BinarySearchTree` `{` ` ` `/* Class containing left and right child` ` ` `of current node and key value*/` ` ` `public` `class` `Node` ` ` `{` ` ` `public` `int` `data;` ` ` `public` `Node left, right;` ` ` `public` `Node(` `int` `item)` ` ` `{` ` ` `data = item;` ` ` `left = right = ` `null` `;` ` ` `}` ` ` `}` ` ` `// Root of BST` ` ` `public` `Node root;` ` ` `// Constructor` ` ` `public` `BinarySearchTree()` ` ` `{` ` ` `root = ` `null` `;` ` ` `}` ` ` `// Returns count of nodes in BST in ` ` ` `// range [low, high]` ` ` `public` `virtual` `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;` ` ` `Console.WriteLine(` `"Count of nodes between ["` `+ l + ` `", "` `+ h + ` `"] is "` `+ tree.getCount(tree.root, l, h));` ` ` `}` `}` ` ` `// This code is contributed by Shrikant13` |

## Javascript

`<script>` `// javascript code to count BST nodes that` `// lie in a given range` ` ` `/*` ` ` `* Class containing left and right child of current node and key value` ` ` `*/` ` ` `class Node {` ` ` `constructor(item) {` ` ` `this` `.data = item;` ` ` `this` `.left =` `this` `.right = ` `null` `;` ` ` `}` ` ` `}` ` ` `// Root of BST` ` ` `var` `root = ` `null` `;` ` ` ` ` `// Returns count of nodes in BST in` ` ` `// range [low, high]` ` ` `function` `getCount( node , low , 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` ` ` ` ` ` ` `root = ` `new` `Node(10);` ` ` `root.left = ` `new` `Node(5);` ` ` `root.right = ` `new` `Node(50);` ` ` `root.left.left = ` `new` `Node(1);` ` ` `root.right.left = ` `new` `Node(40);` ` ` `root.right.right = ` `new` `Node(100);` ` ` `/*` ` ` `* Let us constructed BST shown in above example 10 / \ 5 50 / / \ 1 40 100` ` ` `*/` ` ` `var` `l = 5;` ` ` `var` `h = 45;` ` ` `document.write(` `"Count of nodes between ["` `+ l + ` `", "` `+ h + ` `"] is "` `+ getCount(root, l, h));` `// This code contributed by aashish1995` `</script>` |

**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.

https://youtu.be/jfk

-uX_xKK4?list=PLqM7alHXFySHCXD7r1J0ky9Zg_GBB1dbk

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.

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