Given an undirected graph, how to check if there is a cycle in the graph? For example, the following graph has a cycle 1-0-2-1.
We have discussed cycle detection for the directed graph. We have also discussed a union-find algorithm for cycle detection in undirected graphs.. The time complexity of the union-find algorithm is O(ELogV). Like directed graphs, we can use DFS to detect a cycle in an undirected graph in O(V+E) time. We have discussed DFS based solution for cycle detection in an undirected graph.
In this article, the BFS based solution is discussed. We do a BFS traversal of the given graph. For every visited vertex ‘v’, if there is an adjacent ‘u’ such that u is already visited and u is not a parent of v, then there is a cycle in the graph. If we don’t find such an adjacent for any vertex, we say that there is no cycle. The assumption of this approach is that there are no parallel edges between any two vertices.
We use a parent array to keep track of the parent vertex for a vertex so that we do not consider the visited parent as a cycle.
Time Complexity: The program does a simple BFS Traversal of graph and graph is represented using adjacency list. So the time complexity is O(V+E)