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

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.

Hafeezul Kareem

Updated on 26-Oct-2021 18:25:17

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