# Number of nodes greater than a given value in n-ary tree

Given a n-ary tree and a number x, find and return the number of nodes which are greater than x.

Example:

`In the given tree, x = 7`

`Number of nodes greater than x are 4.`

Approach: The idea is maintain a count variable initialize to 0. Traverse the tree and compare root data with x. If root data is greater than x, increment the count variable and recursively call for all its children.

Below is the implementation of idea.

## C++

 `// C++ program to find number of nodes ``// greater than x ``#include ``using` `namespace` `std; ` `// Structure of a node of n-ary tree ``struct` `Node { ``    ``int` `key; ``    ``vector child; ``}; ` `// Utility function to create ``// a new tree node ``Node* newNode(``int` `key) ``{ ``    ``Node* temp = ``new` `Node; ``    ``temp->key = key; ``    ``return` `temp; ``} ` `// Function to find number of nodes ``// greater than x ``int` `nodesGreaterThanX(Node* root, ``int` `x) ``{ ``    ``if` `(root == NULL) ``        ``return` `0; ` `    ``int` `count = 0; ` `    ``// if current root is greater ``    ``// than x increment count ``    ``if` `(root->key > x) ``        ``count++; ` `    ``// Number of children of root ``    ``int` `numChildren = root->child.size(); ` `    ``// recursively calling for every child ``    ``for` `(``int` `i = 0; i < numChildren; i++) { ``        ``Node* child = root->child[i]; ``        ``count += nodesGreaterThanX(child, x); ``    ``} ` `    ``// return the count ``    ``return` `count; ``} ` `// Driver program ``int` `main() ``{ ``    ``/* Let us create below tree ``*         5 ``*         / | \ ``*     1 2 3 ``*     / / \ \ ``*     15 4 5 6 ``*/` `    ``Node* root = newNode(5); ``    ``(root->child).push_back(newNode(1)); ``    ``(root->child).push_back(newNode(2)); ``    ``(root->child).push_back(newNode(3)); ``    ``(root->child[0]->child).push_back(newNode(15)); ``    ``(root->child[1]->child).push_back(newNode(4)); ``    ``(root->child[1]->child).push_back(newNode(5)); ``    ``(root->child[2]->child).push_back(newNode(6)); ` `    ``int` `x = 5; ` `    ``cout << ``"Number of nodes greater than "``        ``<< x << ``" are "``; ``    ``cout << nodesGreaterThanX(root, x) ``        ``<< endl; ` `    ``return` `0; ``} `

## Java

 `// Java program to find number of nodes ``// greater than x ``import` `java.util.*; ` `// Class representing a Node of an N-ary tree ``class` `Node{ ``    ` `    ``int` `key; ``    ``ArrayList child; ` `    ``// Constructor to create a Node ``    ``Node(``int` `val) ``    ``{ ``        ``key = val; ``        ``child = ``new` `ArrayList<>(); ``    ``} ``} ` `class` `GFG{ ` `// Recursive function to find number ``// of nodes greater than x ``public` `static` `int` `nodesGreaterThanX(Node root, ``int` `x) ``{ ``    ``if` `(root == ``null``) ``        ``return` `0``; ` `    ``int` `count = ``0``; ` `    ``// If current root is greater ``    ``// than x increment count ``    ``if` `(root.key > x) ``        ``count++; ` `    ``// Recursively calling for every ``    ``// child of current root ``    ``for``(Node child : root.child) ``    ``{ ``        ``count += nodesGreaterThanX(child, x); ``    ``} ` `    ``// Return the count ``    ``return` `count; ``} ` `// Driver code ``public` `static` `void` `main(String[] args) ``{ ``    ` `    ``/* Let us create below tree ``            ``5 ``        ``/ | \ ``        ``1 2 3 ``        ``/ / \ \ ``    ``15 4 5 6 ``    ``*/``    ` `    ``Node root = ``new` `Node(``5``); ``    ` `    ``root.child.add(``new` `Node(``1``)); ``    ``root.child.add(``new` `Node(``2``)); ``    ``root.child.add(``new` `Node(``3``)); ``    ` `    ``root.child.get(``0``).child.add(``new` `Node(``15``)); ``    ``root.child.get(``1``).child.add(``new` `Node(``4``)); ``    ``root.child.get(``1``).child.add(``new` `Node(``5``)); ``    ``root.child.get(``2``).child.add(``new` `Node(``6``)); ` `    ``int` `x = ``5``; ` `    ``System.out.print(``"Number of nodes greater than "` `+ ``                    ``x + ``" are "``); ``    ``System.out.println(nodesGreaterThanX(root, x)); ``} ``} ` `// This code is contributed by jrishabh99 `

## Python3

 `# Python3 program to find number of nodes ``# greater than x ` `# Structure of a node of n-ary tree ``class` `Node: ``    ``def` `__init__(``self``, data): ``        ``self``.key ``=` `data ``        ``self``.child ``=` `[] ` `# Function to find number of nodes ``# greater than x ``def` `nodesGreaterThanX(root: Node, x: ``int``) ``-``> ``int``: ``    ``if` `root ``is` `None``: ``        ``return` `0` `    ``count ``=` `0` `    ``# if current root is greater ``    ``# than x increment count ``    ``if` `root.key > x: ``        ``count ``+``=` `1` `    ``# Number of children of root ``    ``numChildren ``=` `len``(root.child) ` `    ``# recursively calling for every child ``    ``for` `i ``in` `range``(numChildren): ``        ``child ``=` `root.child[i] ``        ``count ``+``=` `nodesGreaterThanX(child, x) ` `    ``# return the count ``    ``return` `count ` `# Driver Code ``if` `__name__ ``=``=` `"__main__"``: ` `    ``ans ``=` `0``    ``k ``=` `25` `    ``# Let us create below tree ``    ``# 5 ``    ``#         / | \ ``    ``# 1 2 3 ``    ``#     / / \ \ ``    ``# 15 4 5 6 ` `    ``root ``=` `Node(``5``) ``    ``(root.child).append(Node(``1``)) ``    ``(root.child).append(Node(``2``)) ``    ``(root.child).append(Node(``3``)) ``    ``(root.child[``0``].child).append(Node(``15``)) ``    ``(root.child[``1``].child).append(Node(``4``)) ``    ``(root.child[``1``].child).append(Node(``5``)) ``    ``(root.child[``2``].child).append(Node(``6``)) ` `    ``x ``=` `5` `    ``print``(``"Number of nodes greater than % d are % d"` `%``        ``(x, nodesGreaterThanX(root, x))) ` `# This code is contributed by ``# sanjeev2552 `

## C#

 `// C# program to find number of nodes ``// greater than x ``using` `System;``using` `System.Collections.Generic;`` ` `// Class representing a Node of an N-ary tree ``public` `class` `Node``{ ``    ``public` `int` `key; ``    ``public` `List child; ``    ` `    ``// Constructor to create a Node ``    ``public` `Node(``int` `val) ``    ``{ ``        ``key = val; ``        ``child = ``new` `List(); ``    ``} ``} ` `class` `GFG{ ` `// Recursive function to find number ``// of nodes greater than x ``public` `static` `int` `nodesGreaterThanX(Node root, ``int` `x) ``{ ``    ``if` `(root == ``null``) ``        ``return` `0; ` `    ``int` `count = 0; ` `    ``// If current root is greater ``    ``// than x increment count ``    ``if` `(root.key > x) ``        ``count++; ` `    ``// Recursively calling for every ``    ``// child of current root ``    ``foreach``(Node child ``in` `root.child) ``    ``{ ``        ``count += nodesGreaterThanX(child, x); ``    ``} ` `    ``// Return the count ``    ``return` `count; ``} ` `// Driver code ``public` `static` `void` `Main(String[] args) ``{ ``    ` `    ``/* Let us create below tree ``          ``5 ``        ``/ | \ ``       ``1  2  3 ``      ``/  / \  \ ``    ``15  4   5  6 ``    ``*/``    ``Node root = ``new` `Node(5); ``    ` `    ``root.child.Add(``new` `Node(1)); ``    ``root.child.Add(``new` `Node(2)); ``    ``root.child.Add(``new` `Node(3)); ``    ` `    ``root.child[0].child.Add(``new` `Node(15)); ``    ``root.child[1].child.Add(``new` `Node(4)); ``    ``root.child[1].child.Add(``new` `Node(5)); ``    ``root.child[2].child.Add(``new` `Node(6)); ` `    ``int` `x = 5; ` `    ``Console.Write(``"Number of nodes greater than "` `+ ``                  ``x + ``" are "``); ``    ``Console.WriteLine(nodesGreaterThanX(root, x)); ``} ``} ` `// This code is contributed by Amit Katiyar`

## Javascript

 `// JavaScript program to find number of nodes ``// greater than x ` `// Structure of a node of n-ary tree ``class Node { ``    ``constructor(key) {``        ``this``.key = key; ``        ``this``.child = []; ``    ``}``}` `// Utility function to create ``// a new tree node ``function` `newNode(key) {``    ``return` `new` `Node(key);``}` `// Function to find number of nodes ``// greater than x ``function` `nodesGreaterThanX(root, x) { ``    ``if` `(root == ``null``) ``        ``return` `0; ` `    ``let count = 0; ` `    ``// if current root is greater ``    ``// than x increment count ``    ``if` `(root.key > x) ``        ``count++; ` `    ``// Number of children of root ``    ``const numChildren = root.child.length; ` `    ``// recursively calling for every child ``    ``for` `(let i = 0; i < numChildren; i++) { ``        ``const child = root.child[i]; ``        ``count += nodesGreaterThanX(child, x); ``    ``} ` `    ``// return the count ``    ``return` `count; ``} ` `// Driver program ` `    ``/* Let us create below tree ``    ``*         5 ``    ``*         / | \ ``    ``*     1 2 3 ``    ``*     / / \ \ ``    ``*     15 4 5 6 ``    ``*/` `    ``const root = newNode(5); ``    ``root.child.push(newNode(1)); ``    ``root.child.push(newNode(2)); ``    ``root.child.push(newNode(3)); ``    ``root.child[0].child.push(newNode(15)); ``    ``root.child[1].child.push(newNode(4)); ``    ``root.child[1].child.push(newNode(5)); ``    ``root.child[2].child.push(newNode(6)); ` `    ``const x = 5; ` `    ``console.log(``"Number of nodes greater than "` `+ x + ``" are "` `+ nodesGreaterThanX(root, x));`

Output
`Number of nodes greater than 5 are 2`

Time complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(h) where h is the height of the binary tree.

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