Number of leaf nodes in the subtree of every node of an n-ary tree

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Given an N-ary tree, print the number of leaf nodes in the subtree of every node.

Examples:

Input:      
            1 
          /    \
         2      3
             /  |   \
            4   5    6
Output:
The node 1 has 4 leaf nodes
The node 2 has 1 leaf nodes
The node 3 has 3 leaf nodes
The node 4 has 1 leaf nodes
The node 5 has 1 leaf nodes
The node 6 has 1 leaf nodes

Approach: The idea is to perform DFS traversal on the given tree and for every node keep an array leaf[] to store the count of leaf nodes in the subtree below it.
Now, while recurring down the tree, if a leaf node is found set its leaf[i] value to 1 and return back in upward direction. Now, every time while returning back from the function call to upward, add the leaf nodes of the node below it.

Once, the DFS traversal is completed we will have the count of leaf nodes in the array leaf[].

Below is the implementation of the above approach:

C++

// C++ program to print the number of
// leaf nodes of every node
#include <bits/stdc++.h>
using namespace std;
 
// Function to insert edges of tree
void insert(int x, int y, vector<int> adjacency[])
{
    adjacency[x].push_back(y);
}
 
// Function to run DFS on a tree
void dfs(int node, int leaf[], int vis[],
         vector<int> adjacency[])
{
    leaf[node] = 0;
    vis[node] = 1;
 
    // iterate on all the nodes
    // connected to node
    for (auto it : adjacency[node]) {
 
        // If not visited
        if (!vis[it]) {
            dfs(it, leaf, vis, adjacency);
            leaf[node] += leaf[it];
        }
    }
 
    if (!adjacency[node].size())
        leaf[node] = 1;
}
 
// Function to print number of
// leaf nodes of a node
void printLeaf(int n, int leaf[])
{
    // Function to print leaf nodes
    for (int i = 1; i <= n; i++) {
        cout << "The node " << i << " has "
             << leaf[i] << " leaf nodes\n";
    }
}
 
// Driver Code
int main()
{
    // Given N-ary Tree
 
    /*     1 
         /   \
        2     3
            / | \
            4 5 6 */
 
    int N = 6; // no of nodes
    vector<int> adjacency[N + 1]; // adjacency list for tree
 
    insert(1, 2, adjacency);
    insert(1, 3, adjacency);
    insert(3, 4, adjacency);
    insert(3, 5, adjacency);
    insert(3, 6, adjacency);
 
    int leaf[N + 1]; // Store count of leaf in subtree of i
    int vis[N + 1] = { 0 }; // mark nodes visited
 
    dfs(1, leaf, vis, adjacency);
 
    printLeaf(N, leaf);
 
    return 0;
}

Java

// Java program to print the number of
// leaf nodes of every node
import java.util.*;
 
class GFG
{
static Vector<Vector<Integer>> adjacency = new
       Vector<Vector<Integer>>();
 
// Function to insert edges of tree
static void insert(int x, int y)
{
    adjacency.get(x).add(y);
}
 
// Function to run DFS on a tree
static void dfs(int node, int leaf[], int vis[])
{
    leaf[node] = 0;
    vis[node] = 1;
 
    // iterate on all the nodes
    // connected to node
    for (int i = 0; i < adjacency.get(node).size(); i++)
    {
        int it = adjacency.get(node).get(i);
         
        // If not visited
        if (vis[it] == 0) 
        {
            dfs(it, leaf, vis);
            leaf[node] += leaf[it];
        }
    }
 
    if (adjacency.get(node).size() == 0)
        leaf[node] = 1;
}
 
// Function to print number of
// leaf nodes of a node
static void printLeaf(int n, int leaf[])
{
    // Function to print leaf nodes
    for (int i = 1; i <= n; i++) 
    {
        System.out.print( "The node " + i + " has " + 
                          leaf[i] + " leaf nodes\n");
    }
}
 
// Driver Code
public static void main(String args[])
{
    // Given N-ary Tree
 
    /*     1 
        / \
        2     3
            / | \
            4 5 6 */
 
    int N = 6; // no of nodes
     
    for(int i = 0; i <= N; i++)
    adjacency.add(new Vector<Integer>());
     
    insert(1, 2);
    insert(1, 3);
    insert(3, 4);
    insert(3, 5);
    insert(3, 6);
 
    // Store count of leaf in subtree of i
    int leaf[] = new int[N + 1]; 
     
    // mark nodes visited
    int vis[] = new int[N + 1] ; 
 
    dfs(1, leaf, vis);
 
    printLeaf(N, leaf);
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program to print the number of
# leaf nodes of every node
adjacency = [[] for i in range(100)]
 
# Function to insert edges of tree
def insert(x, y):
    adjacency[x].append(y)
 
# Function to run DFS on a tree
def dfs(node, leaf, vis):
 
    leaf[node] = 0
    vis[node] = 1
 
    # iterate on all the nodes
    # connected to node
    for it in adjacency[node]:
 
        # If not visited
        if (vis[it] == False):
            dfs(it, leaf, vis)
            leaf[node] += leaf[it]
 
    if (len(adjacency[node]) == 0):
        leaf[node] = 1
 
# Function to print number of
# leaf nodes of a node
def printLeaf(n, leaf):
     
    # Function to print leaf nodes
    for i in range(1, n + 1):
        print("The node", i, "has",  
               leaf[i], "leaf nodes")
 
# Driver Code
 
# Given N-ary Tree
'''
/*     1
    / \
    2     3
        / | \
        4 5 6 '''
 
N = 6 # no of nodes
# adjacency list for tree
 
insert(1, 2)
insert(1, 3)
insert(3, 4)
insert(3, 5)
insert(3, 6)
 
# Store count of leaf in subtree of i
leaf = [0 for i in range(N + 1)] 
 
# mark nodes visited
vis = [0 for i in range(N + 1)] 
 
dfs(1, leaf, vis)
 
printLeaf(N, leaf)
 
# This code is contributed by Mohit Kumar

C#

// C# program to print the number of
// leaf nodes of every node
using System;
using System.Collections.Generic;
 
class GFG
{
static List<List<int>> adjacency = new
       List<List<int>>();
  
// Function to insert edges of tree
static void insert(int x, int y)
{
    adjacency[x].Add(y);
}
  
// Function to run DFS on a tree
static void dfs(int node, int []leaf, int []vis)
{
    leaf[node] = 0;
    vis[node] = 1;
  
    // iterate on all the nodes
    // connected to node
    for (int i = 0; i < adjacency[node].Count; i++)
    {
        int it = adjacency[node][i];
          
        // If not visited
        if (vis[it] == 0) 
        {
            dfs(it, leaf, vis);
            leaf[node] += leaf[it];
        }
    }
  
    if (adjacency[node].Count == 0)
        leaf[node] = 1;
}
  
// Function to print number of
// leaf nodes of a node
static void printLeaf(int n, int []leaf)
{
    // Function to print leaf nodes
    for (int i = 1; i <= n; i++) 
    {
        Console.Write( "The node " + i + " has " + 
                          leaf[i] + " leaf nodes\n");
    }
}
  
// Driver Code
public static void Main(String []args)
{
    // Given N-ary Tree
  
    /*     1 
        / \
        2     3
            / | \
            4 5 6 */
  
    int N = 6; // no of nodes
      
    for(int i = 0; i <= N; i++)
    adjacency.Add(new List<int>());
      
    insert(1, 2);
    insert(1, 3);
    insert(3, 4);
    insert(3, 5);
    insert(3, 6);
  
    // Store count of leaf in subtree of i
    int []leaf = new int[N + 1]; 
      
    // mark nodes visited
    int []vis = new int[N + 1] ; 
  
    dfs(1, leaf, vis);
  
    printLeaf(N, leaf);
}
}
 
// This code contributed by Rajput-Ji

Javascript

<script>
 
// JavaScript program to print the number of
// leaf nodes of every node
 
let adjacency = [];
 
// Function to insert edges of tree
function insert(x,y)
{
    adjacency[x].push(y);
}
 
// Function to run DFS on a tree
function dfs(node,leaf,vis)
{
    leaf[node] = 0;
    vis[node] = 1;
   
    // iterate on all the nodes
    // connected to node
    for (let i = 0; i < adjacency[node].length; i++)
    {
        let it = adjacency[node][i];
           
        // If not visited
        if (vis[it] == 0) 
        {
            dfs(it, leaf, vis);
            leaf[node] += leaf[it];
        }
    }
   
    if (adjacency[node].length == 0)
        leaf[node] = 1;
}
 
// Function to print number of
// leaf nodes of a node
function printLeaf(n,leaf)
{
    // Function to print leaf nodes
    for (let i = 1; i <= n; i++) 
    {
        document.write( "The node " + i + " has " + 
                          leaf[i] + " leaf nodes<br>");
    }
}
 
// Driver Code
 
// Given N-ary Tree
 
/*     1 
        / \
        2     3
            / | \
            4 5 6 */
 
let N = 6; // no of nodes
 
for(let i = 0; i <= N; i++)
    adjacency.push([]);
 
insert(1, 2);
insert(1, 3);
insert(3, 4);
insert(3, 5);
insert(3, 6);
 
// Store count of leaf in subtree of i
let leaf = new Array(N + 1); 
for(let i=0;i<leaf.length;i++)
{
    leaf[i]=0;
}
// mark nodes visited
let vis = new Array(N + 1) ; 
for(let i=0;i<vis.length;i++)
{
    vis[i]=0;
}
 
dfs(1, leaf, vis);
 
printLeaf(N, leaf);
 
     
 
// This code is contributed by patel2127
 
</script>

Output

The node 1 has 4 leaf nodes
The node 2 has 1 leaf nodes
The node 3 has 3 leaf nodes
The node 4 has 1 leaf nodes
The node 5 has 1 leaf nodes
The node 6 has 1 leaf nodes

Complexity Analysis:

Time Complexity: O(N), as we are traversing the tree using DFS(recursion), where N is the number of nodes in the tree.
Auxiliary Space: O(N), as we are using extra space for the tree. Where N is the number of nodes in the tree.

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