Find all paths in graph from A to N

Question

I'm trying to port the following example python code to Java:

def find_all_paths(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return [path]
    if not graph.has_key(start):
        return []
    paths = []
    for node in graph[start]:
        if node not in path:
            newpaths = find_all_paths(graph, node, end, path)
            for newpath in newpaths:
                paths.append(newpath)
    return paths

The problem being, the base case to stop recursing:

if start == end:
    return [path]

It doesn't support my requirement of allowing both A and N to be the same node.

For example:

If I have the following digraph :

A -> [B, C], 
B -> [C, E], 
C -> [D, A]

And I want all paths between A and A, I should have the result:

A -> B -> C -> A

The above python code will just give me:

A

Answer 1

A path from A to A must go through a neighbor of A. Thus, one way to implement this is to enumerate all of the outward neighbors:

[[["A"]+y for y in find_all_paths(G,x,"A")] for x in graph["A"]]

For your graph, the result should be

[[['A', 'B', 'C', 'A']], [['A', 'C', 'A']]]