# Count BST nodes that lie in a given range

• Difficulty Level : Easy
• Last Updated : 22 Sep, 2022

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

Examples:

Input: Range: [5, 45]

10
/     \
5       50
/        /    \
1      40   100
Output:  3
Explanation: There are three nodes in range, 5, 10 and 40

Input: Range: [10, 100]

10
/    \
5      50
/       /    \
1     40   100
Output:  4

Approach:

Traverse the given binary search tree starting from root. For every node 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.

Follow the below steps to Implement the idea:

• Traverse the tree starting from the root.
• If root->data is equal to high and root->data is equal to low return 1.
• If root->data <= high and root->data >= low then return 1 + count on left of root + count on right of root.
• Else if root->data < low return count on right of root.
• Else if root->data > high return count on right of root.

Below is the implementation of the above approach.

## C++

 `// C++ program to count BST nodes within a given range``#include``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``# within 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

 ``

Output

`Count of nodes between [5, 45] is 3`

Time complexity: O(H + k) where h is the height of BST and k is the number of nodes in the given range.
Auxiliary Space: O(n)

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