Print all neighbour nodes within distance K
Given a graph of N nodes, E edges, a node X and a distance K. The task is to print all the nodes within the distance K from X.
Input:
Output: 4 5 2 7 3
Neigbour nodes within distance 2 of node 4 are: 4 5 2 7 3
Approach:
To print all the nodes that are at distance K or less than K. We can do it by applying dfs variation, that takes K node from where we have to print the distance until distance K.
dfs(K, node, -1, tree)
Here -1 indicates node parent.
This recursive function basically prints the node and then calls the dfs(K-1, neighbour of node, node, tree).
Base condition is K>0.
Below is the implementation of the above approach:
C++
// C++ program to print// the nearest K neighbour// nodes (including itself)#include <bits/stdc++.h>using namespace std;// Structure of an edgestruct arr { int from, to;};// Recursive function to print// the neighbor nodes of a node// until K distancevoid dfs(int k, int node, int parent, const vector<vector<int> >& tree){ // Base condition if (k < 0) return; // Print the node cout << node << ' '; // Traverse the connected // nodes/adjacency list for (int i : tree[node]) { if (i != parent) { // node i becomes the parent // of its child node dfs(k - 1, i, node, tree); } }}// Function to print nodes under// distance kvoid print_under_dis_K(struct arr graph[], int node, int k, int v, int e){ // To make graph with // the given edges vector<vector<int> > tree(v + 1, vector<int>()); for (int i = 0; i < e; i++) { int from = graph[i].from; int to = graph[i].to; tree[from].push_back(to); tree[to].push_back(from); } dfs(k, node, -1, tree);}// Driver Codeint main(){ // Number of vertex and edges int v = 7, e = 6; // Given edges struct arr graph[v + 1] = { { 2, 1 }, { 2, 5 }, { 5, 4 }, { 5, 7 }, { 4, 3 }, { 7, 6 } }; // k is the required distance // upto which are neighbor // nodes should get printed int node = 4, k = 2; // function calling print_under_dis_K(graph, node, k, v, e); return 0;} |
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Java
// Java program to print // the nearest K neighbour // nodes (including itself) import java.util.*;@SuppressWarnings("unchecked") class GFG{ // Structure of an edge public static class arr { public int from, to; public arr(int from, int to) { this.from = from; this.to = to; }}; // Recursive function to print // the neighbor nodes of a node // until K distance static void dfs(int k, int node, int parent, ArrayList []tree) { // Base condition if (k < 0) return; // Print the node System.out.print(node + " "); ArrayList tmp = (ArrayList)tree[node]; // Traverse the connected // nodes/adjacency list for(int i : (ArrayList<Integer>)tmp) { if (i != parent) { // Node i becomes the parent // of its child node dfs(k - 1, i, node, tree); } } } // Function to print nodes under // distance k static void print_under_dis_K(arr []graph, int node, int k, int v, int e) { // To make graph with // the given edges ArrayList []tree = new ArrayList[v + 1]; for(int i = 0; i < v + 1; i++) { tree[i] = new ArrayList(); } for(int i = 0; i < e; i++) { int from = graph[i].from; int to = graph[i].to; tree[from].add(to); tree[to].add(from); } dfs(k, node, -1, tree); } // Driver Codepublic static void main(String[] args){ // Number of vertex and edges int v = 7, e = 6; // Given edges arr []graph = { new arr(2, 1), new arr(2, 5), new arr(5, 4), new arr(5, 7), new arr(4, 3), new arr(7, 6) }; // k is the required distance // upto which are neighbor // nodes should get printed int node = 4, k = 2; // Function calling print_under_dis_K(graph, node, k, v, e); }}// This code is contributed by pratham76 |
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Python3
# Python3 program to print# the nearest K neighbour# nodes (including itself)tree = [[] for i in range(100)]# Recursive function to print# the neighbor nodes of a node# until K distancedef dfs(k, node, parent): # Base condition if (k < 0): return # Print the node print(node, end = " ") # Traverse the connected # nodes/adjacency list for i in tree[node]: if (i != parent): # node i becomes the parent # of its child node dfs(k - 1, i, node)# Function to print nodes under# distance kdef print_under_dis_K(graph, node, k, v, e): for i in range(e): fro = graph[i][0] to = graph[i][1] tree[fro].append(to) tree[to].append(fro) dfs(k, node, -1)# Driver Code# Number of vertex and edgesv = 7e = 6# Given edgesgraph = [[ 2, 1 ], [ 2, 5 ], [ 5, 4 ], [ 5, 7 ], [ 4, 3 ], [ 7, 6 ]] # k is the required distance# upto which are neighbor# nodes should get prednode = 4k = 2# function callingprint_under_dis_K(graph, node, k, v, e)# This code is contributed by Mohit Kumar |
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C#
// C# program to print // the nearest K neighbour // nodes (including itself) using System;using System.Collections;class GFG{ // Structure of an edge public class arr { public int from, to; public arr(int from, int to) { this.from = from; this.to = to; }}; // Recursive function to print // the neighbor nodes of a node // until K distance static void dfs(int k, int node, int parent, ArrayList []tree) { // Base condition if (k < 0) return; // Print the node Console.Write(node+" "); ArrayList tmp = (ArrayList)tree[node]; // Traverse the connected // nodes/adjacency list foreach (int i in tmp) { if (i != parent) { // node i becomes the parent // of its child node dfs(k - 1, i, node, tree); } } } // Function to print nodes under // distance k static void print_under_dis_K(arr []graph, int node, int k, int v, int e) { // To make graph with // the given edges ArrayList []tree = new ArrayList[v + 1]; for(int i = 0; i < v + 1; i++) { tree[i] = new ArrayList(); } for (int i = 0; i < e; i++) { int from = graph[i].from; int to = graph[i].to; tree[from].Add(to); tree[to].Add(from); } dfs(k, node, -1, tree); } // Driver Code public static void Main(string[] args) { // Number of vertex and edges int v = 7, e = 6; // Given edges arr []graph = { new arr( 2, 1 ), new arr( 2, 5 ), new arr( 5, 4 ), new arr( 5, 7 ), new arr( 4, 3 ), new arr( 7, 6 ) }; // k is the required distance // upto which are neighbor // nodes should get printed int node = 4, k = 2; // function calling print_under_dis_K(graph, node, k, v, e); }}// This code is contributed by rutvik_56 |
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Output:
4 5 2 7 3
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