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 edge struct arr { int from, to; }; // Recursive function to print // the neighbor nodes of a node // until K distance void 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 k void 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 Code int 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; } |
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 distance def 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 k def 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 edges v = 7e = 6 # Given edges graph = [[ 2, 1 ], [ 2, 5 ], [ 5, 4 ], [ 5, 7 ], [ 4, 3 ], [ 7, 6 ]] # k is the required distance # upto which are neighbor # nodes should get pred node = 4k = 2 # function calling print_under_dis_K(graph, node, k, v, e) # This code is contributed by Mohit Kumar |
4 5 2 7 3
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