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Print all leaf nodes of an n-ary tree using DFS
  • Difficulty Level : Basic
  • Last Updated : 26 Nov, 2020

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

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// C# implementation of the approach
using System.Collections;
using System.Collections.Generic;
using System;
 
class GFG{
     
// Pair class
class pair
{
    public int first, second;
    public pair(int a, int b)
    {
        first = a;
        second = b;
    }
}
 
// Function to perform DFS on the tree
static void dfs(ArrayList t, int node,
                             int parent)
{
    int flag = 1;
     
    // Iterating the children of current node
    for(int i = 0;
            i < ((ArrayList)t[node]).Count;
            i++)
    {
        int ir = (int)((ArrayList)t[node])[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)
        Console.Write( node + " ");
}
 
// Driver code
public static void Main(string []args)
{
     
    // Adjacency list
    ArrayList t = new ArrayList();
 
    // 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;
     
    for(int i = 0; i < 1005; i++)
    {
        t.Add(new ArrayList());
    }
 
    // Create the tree
    for(int i = 0; i < cnt; i++)
    {
        ((ArrayList)t[edges[i].first]).Add(
            edges[i].second);
        ((ArrayList)t[edges[i].second]).Add(
            edges[i].first);
    }
 
    // Function call
    dfs(t, 1, 0);
}
}
 
// This code is contributed by rutvik_56

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Output: 

4 5 8 7

 

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