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

Given an n-ary tree and a number, we have to count the number of nodes greater than the given number. Let's see an example.

Input

tree = [[4], [1, 2], [3, 5]]
n = 2

Output

3

There are 3 nodes with values that are greater than n.

Algorithm

• Initialise the n-ary tree.

• Initialise the count to 0.

• Increment the count by 1 when you find a node whose value is greater than n.

• Get the children of the current node.

• Iterate over all the children and recursively call the same function to count nodes.

• Return the count.

Implementation

Following is the implementation of the above algorithm in C++

#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
vector<Node*> child;
};
Node* getNewNode(int data) {
Node* temp = new Node;
temp->data = data;
return temp;
}
int getGreaterElementsCount(Node* root, int n) {
if (root == NULL)
return 0;
int count = 0;
if (root->data > n) {
count++;
}
int nodeChildrenCount = root->child.size();
for (int i = 0; i < nodeChildrenCount; i++) {
Node* child = root->child[i];
count += getGreaterElementsCount(child, n);
}
return count;
}
int main() {
Node* root = getNewNode(1);
(root->child).push_back(getNewNode(2));
(root->child).push_back(getNewNode(3));
(root->child).push_back(getNewNode(4));
(root->child[0]->child).push_back(getNewNode(5));
(root->child[0]->child).push_back(getNewNode(5));
(root->child[1]->child).push_back(getNewNode(6));
(root->child[1]->child).push_back(getNewNode(6));
(root->child[1]->child).push_back(getNewNode(7));
(root->child[2]->child).push_back(getNewNode(8));
(root->child[2]->child).push_back(getNewNode(8));
(root->child[2]->child).push_back(getNewNode(9));
int n = 2;
cout << getGreaterElementsCount(root, n) << endl;
return 0;
}

Output

If you run the above code, then you will get the following result.

10