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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to perform DFS on the treevoid 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 codeint 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;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{ // Pair classstatic class pair{ int first,second; pair(int a, int b) { first = a; second = b; }}// Function to perform DFS on the treestatic 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 codepublic 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 |
Python3
# Python3 implementation of the approacht = [[] for i in range(1005)]# Function to perform DFS on the treedef 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 edgesedges = [[ 1, 2 ], [ 1, 3 ], [ 2, 4 ], [ 3, 5 ], [ 3, 6 ], [ 3, 7 ], [ 6, 8 ]]# Count of edgescnt = len(edges)# Number of nodesnode = cnt + 1# Create the treefor i in range(cnt): t[edges[i][0]].append(edges[i][1]) t[edges[i][1]].append(edges[i][0])# Function calldfs(1, 0)# This code is contributed by Mohit Kumar |
C#
// 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 |
Javascript
<script>// Javascript implementation of the approach// Function to perform DFS on the treefunction dfs(t, node, parent){ let flag = 1; // Iterating the children of current node for(let i = 0; i < t[node].length; i++) { let ir = 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) document.write( node + " ");}// Driver code// Adjacency listlet t = []// List of all edgeslet edges = [ [ 1, 2 ], [ 1, 3 ], [ 2, 4 ], [ 3, 5 ], [ 3, 6 ], [ 3, 7 ], [ 6, 8 ] ];// Count of edgeslet cnt = edges.length;// Number of nodeslet node = cnt + 1; for(let i = 0; i < 1005; i++){ t.push([]);}// Create the treefor(let i = 0; i < cnt; i++){ t[edges[i][0]].push(edges[i][1]) t[edges[i][1]].push(edges[i][0])}// Function calldfs(t, 1, 0);// This code is contributed by patel2127</script> |
Output:
4 5 8 7
Time Complexity: O(N), where N is the number of nodes in the graph.
Auxiliary Space: O(N)
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