Why DFS and not BFS for finding cycle in graphs

Question

Predominantly DFS is used to find a cycle in graphs and not BFS. Any reasons? Both can find if a node has already been visited while traversing the tree/graph.

Answer 1

A BFS could be reasonable if the graph is undirected (be my guest at showing an efficient algorithm using BFS that would report the cycles in a directed graph!), where each "cross edge" defines a cycle. If the cross edge is {v1, v2}, and the root (in the BFS tree) that contains those nodes is r, then the cycle is r ~ v1 - v2 ~ r (~ is a path, - a single edge), which can be reported almost as easily as in DFS.

The only reason to use a BFS would be if you know your (undirected) graph is going to have long paths and small path cover (in other words, deep and narrow). In that case, BFS would require proportionally less memory for its queue than DFS' stack (both still linear of course).

In all other cases, DFS is clearly the winner. It works on both directed and undirected graphs, and it is trivial to report the cycles - just concat any back edge to the path from the ancestor to the descendant, and you get the cycle. All in all, much better and practical than BFS for this problem.