# Print all leaf nodes of an n-ary tree using DFS

Given an array **edge[][2]** where **(edge[i][0], edge[i][1])** defines an edge in the n-ary tree, the task is to print all the leaf nodes of the given tree using.

**Examples:**

Input:edge[][] = {{1, 2}, {1, 3}, {2, 4}, {2, 5}, {3, 6}}Output:4 5 6 1 / \ 2 3 / \ \ 4 5 6Input:edge[][] = {{1, 5}, {1, 7}, {5, 6}}Output:6 7

**Approach:** DFS can be used to traverse the complete tree. We will keep track of parent while traversing to avoid the visited node array. Initially for every node we can set a flag and if the node have at least one child (i.e. non-leaf node) then we will reset the flag. The nodes with no children are the leaf nodes.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to perform DFS on the tree ` `void` `dfs(list<` `int` `> t[], ` `int` `node, ` `int` `parent) ` `{ ` ` ` `int` `flag = 1; ` ` ` ` ` `// Iterating the children of current node ` ` ` `for` `(` `auto` `ir : t[node]) { ` ` ` ` ` `// There is at least a child ` ` ` `// of the current node ` ` ` `if` `(ir != parent) { ` ` ` `flag = 0; ` ` ` `dfs(t, ir, node); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Current node is connected to only ` ` ` `// its parent i.e. it is a leaf node ` ` ` `if` `(flag == 1) ` ` ` `cout << node << ` `" "` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `// Adjacency list ` ` ` `list<` `int` `> t[1005]; ` ` ` ` ` `// List of all edges ` ` ` `pair<` `int` `, ` `int` `> edges[] = { { 1, 2 }, ` ` ` `{ 1, 3 }, ` ` ` `{ 2, 4 }, ` ` ` `{ 3, 5 }, ` ` ` `{ 3, 6 }, ` ` ` `{ 3, 7 }, ` ` ` `{ 6, 8 } }; ` ` ` ` ` `// Count of edges ` ` ` `int` `cnt = ` `sizeof` `(edges) / ` `sizeof` `(edges[0]); ` ` ` ` ` `// Number of nodes ` ` ` `int` `node = cnt + 1; ` ` ` ` ` `// Create the tree ` ` ` `for` `(` `int` `i = 0; i < cnt; i++) { ` ` ` `t[edges[i].first].push_back(edges[i].second); ` ` ` `t[edges[i].second].push_back(edges[i].first); ` ` ` `} ` ` ` ` ` `// Function call ` ` ` `dfs(t, 1, 0); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python3 implementation of the approach

t = [[] for i in range(1005)]

# Function to perform DFS on the tree

def dfs(node, parent):

flag = 1

# Iterating the children of current node

for ir in t[node]:

# There is at least a child

# of the current node

if (ir != parent):

flag = 0

dfs(ir, node)

# Current node is connected to only

# its parent i.e. it is a leaf node

if (flag == 1):

print(node, end = ” “)

# Driver code

# List of all edges

edges = [[ 1, 2 ],

[ 1, 3 ],

[ 2, 4 ],

[ 3, 5 ],

[ 3, 6 ],

[ 3, 7 ],

[ 6, 8 ]]

# Count of edges

cnt = len(edges)

# Number of nodes

node = cnt + 1

# Create the tree

for i in range(cnt):

t[edges[i][0]].append(edges[i][1])

t[edges[i][1]].append(edges[i][0])

# Function call

dfs(1, 0)

# This code is contributed by Mohit Kumar

**Output:**

4 5 8 7

## Recommended Posts:

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