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.

// CPP program to find number of nodes // greater than x #include <bits/stdc++.h> using namespace std; // Structure of a node of n-ary tree struct Node { int key; vector<Node*> 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 nuber of nodes // gretaer 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; }

Output:

Number of nodes greater than 5 are 2

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