# 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   6

Input: edge[][] = {{1, 5}, {1, 7}, {5, 6}}
Output: 6 7
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` `} `

## 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 `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `// Pair class ` `static` `class` `pair ` `{ ` `    ``int` `first,second; ` `    ``pair(``int` `a, ``int` `b) ` `    ``{ ` `        ``first = a; ` `        ``second = b; ` `    ``} ` `} ` ` `  `// Function to perform DFS on the tree ` `static` `void` `dfs(Vector t, ``int` `node, ``int` `parent) ` `{ ` `    ``int` `flag = ``1``; ` `     `  `    ``// Iterating the children of current node ` `    ``for` `(``int` `i = ``0``; i < ((Vector)t.get(node)).size(); i++)  ` `    ``{ ` `        ``int` `ir = (``int``)((Vector)t.get(node)).get(i); ` `         `  `        ``// 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``) ` `        ``System.out.print( node + ``" "``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``// Adjacency list ` `    ``Vector t = ``new` `Vector(); ` ` `  `    ``// List of all edges ` `    ``pair edges[] = { ``new` `pair( ``1``, ``2` `), ` `                    ``new` `pair( ``1``, ``3` `), ` `                    ``new` `pair( ``2``, ``4` `), ` `                    ``new` `pair( ``3``, ``5` `), ` `                    ``new` `pair( ``3``, ``6` `), ` `                    ``new` `pair( ``3``, ``7` `), ` `                    ``new` `pair( ``6``, ``8` `) }; ` ` `  `    ``// Count of edges ` `    ``int` `cnt = edges.length; ` ` `  `    ``// Number of nodes ` `    ``int` `node = cnt + ``1``; ` `     `  `    ``for``(``int` `i = ``0``; i < ``1005``; i++) ` `    ``{ ` `        ``t.add(``new` `Vector()); ` `    ``} ` ` `  `    ``// Create the tree ` `    ``for` `(``int` `i = ``0``; i < cnt; i++) ` `    ``{ ` `        ``((Vector)t.get(edges[i].first)).add(edges[i].second); ` `        ``((Vector)t.get(edges[i].second)).add(edges[i].first); ` `    ``} ` ` `  `    ``// Function call ` `    ``dfs(t, ``1``, ``0``); ` `} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

Output:

```4 5 8 7
```

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