Detect cycle in the graph using degrees of nodes of graph

Given a graph, the task is to detect a cycle in the graph using degrees of the nodes in the graph and print all the nodes that are involved in any of the cycles. If there is no cycle in the graph then print -1.



Output: 0 1 2

Approach: Recursively remove all vertices of degree 1. This can be done efficiently by storing a map of vertices to their degrees.
Initially, traverse the map and store all the vertices with degree = 1 in a queue. Traverse the queue as long as it is not empty. For each node in the queue, mark it as visited, and iterate through all the nodes that are connected to it (using the adjacency list), and decrement the degree of each of those nodes by one in the map. Add all nodes whose degree becomes equal to one to the queue. At the end of this algorithm, all the nodes that are unvisited are part of the cycle.

Below is the implementation of the above approach:





// Java implementation of the approach
import java.util.*;
// Graph class
class Graph {
    // No. of vertices of graph
    int v;
    // Adjacency List
    LinkedList l[];
    Graph(int v)
        this.v = v;
        this.l = new LinkedList[v];
        for (int i = 0; i < v; i++) {
            l[i] = new LinkedList();
    void addedge(int i, int j)
class GFG {
    // Function to find a cycle in the given graph if exists
    static void findCycle(int n, int e, Graph g)
        // HashMap to store the degree of each node
        HashMap degree = new HashMap();
        for (int i = 0; i < g.l.length; i++)
            degree.put(i, g.l[i].size());
        // Array to track visited nodes
        int visited[] = new int[g.v];
        // Initially all nodes are not visited
        for (int i = 0; i < visited.length; i++)
            visited[i] = 0;
        // Queue to store the nodes of degree 1
        LinkedList q = new LinkedList();
        // Continuously adding those nodes whose
        // degree is 1 to the queue
        while (true) {
            // Adding nodes to queue whose degree is 1
            // and is not visited
            for (int i = 0; i < degree.size(); i++)
                if ((int)degree.get(i) == 1 && visited[i] == 0)
            // If queue becomes empty then get out
            // of the continuous loop
            if (q.isEmpty())
            while (!q.isEmpty()) {
                // Remove the front element from the queue
                int temp = (int)q.remove();
                // Mark the removed element visited
                visited[temp] = 1;
                // Decrement the degree of all those nodes
                // adjacent to removed node
                for (int i = 0; i < g.l[temp].size(); i++) {
                    int value = (int)degree.get((int)g.l[temp].get(i));
                    degree.replace(g.l[temp].get(i), --value);
        int flag = 0;
        // Checking all the nodes which are not visited
        // i.e. they are part of the cycle
        for (int i = 0; i < visited.length; i++)
            if (visited[i] == 0)
                flag = 1;
        if (flag == 0)
        else {
            for (int i = 0; i < visited.length; i++)
                if (visited[i] == 0)
                    System.out.print(i + " ");
    // Driver code
    public static void main(String[] args)
        // No of nodes
        int n = 5;
        // No of edges
        int e = 5;
        Graph g = new Graph(n);
        g.addedge(0, 1);
        g.addedge(0, 2);
        g.addedge(0, 3);
        g.addedge(1, 2);
        g.addedge(3, 4);
        findCycle(n, e, g);



0 1 2

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at to report any issue with the above content.