Number of Leaf Nodes in Subtree of Every Node in N-ary Tree in C++

In this tutorial, we are going to write a program that finds the number of leaf nodes for every node in the n-ary tree.

Given a n-ary tree, we have to find the number of leaf nodes for every subtree. Let's see an example.

Input

N = 8
tree = [[2, 3], [], [4, 5, 6], [7, 8], [], [], [], []]

Output

1->5 2->1 3->4 4->2 5->1 6->1 7->1 8->1

Algorithm

• Initialise the n-ary tree with tree you like.

• Use the DFS to traverse through the tree.

• Maintain an array to store the count of each node leaf nodes count.

• Increment the count of leaf node after the recursive call to DFS.

• Print all node with leaf nodes count.

Implementation

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

#include <bits/stdc++.h>
using namespace std;
void insertNode(int x, int y, vector<int> tree[]) {
   tree[x].push_back(y);
}
void DFS(int node, int leaf[], int visited[], vector<int> tree[]) {
   leaf[node] = 0;
   visited[node] = 1;
   for (auto it : tree[node]) {
      if (!visited[it]) {
         DFS(it, leaf, visited, tree);
         leaf[node] += leaf[it];
      }
   }
   if (!tree[node].size()) {
      leaf[node] = 1;
   }
}
int main() {
   int N = 8;
   vector<int> tree[N + 1];
   insertNode(1, 2, tree);
   insertNode(1, 3, tree);
   insertNode(3, 4, tree);
   insertNode(3, 5, tree);
   insertNode(3, 6, tree);
   insertNode(4, 7, tree);
   insertNode(4, 8, tree);
   int leaf[N + 1];
   int visited[N + 1];
   for (int i = 0; i <= N; i++) {
      visited[i] = 0;
   }
   DFS(1, leaf, visited, tree);
   for (int i = 1; i <= N; i++) {
      cout << i << "->" << leaf[i] << endl;
   }
   return 0;
}

Output

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

1->5
2->1
3->4
4->2
5->1
6->1
7->1
8->1