Number of leaf nodes in the subtree of every node of an 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

Hafeezul Kareem

Published on 03-Jul-2021 08:10:52

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