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

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

**Examples**:

Input: 1 / \ 2 3 / | \ 4 5 6Output: 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 it’s 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 isnert 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; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python3 program to prthe number of

# leaf nodes of every node

adjacency = [[] for i in range(100)]

# Function to isnert 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 prnumber of

# leaf nodes of a node

def printLeaf(n, leaf):

# Function to prleaf 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

**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

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