提问

上图是我的一个练习作业的问题陈述。

For the above problem, this was the idea I came up w/ for my recursive algorithm
Base case = you've hit a node u with no outgoing edges. `return [u]`
If there are outgoing edges from current node u to vertices v;

src_paths = []
for each v
    child_paths = recursivealgorithm(v)
    for each child_path in child_paths:
        src_path = [u] + child_path
        src_paths.append(src_path)
return src_paths

The trouble w/ this idea as defined is that there is no way to stop an infinite loop happening in case of a cycle in the graph. I thought of various ideas to prevent such a loop (such as keeping track of the vertices visited thus far, maintaining a dictionary including the distance of each visited vertex from the original src & somehow using that to tell if there was a cycle, etc.) but they all had the trouble of also causing issues if 2 child_paths ever converged at a vertex (without creating a cycle)

For eg

上面的图片没有循环，但是如果我写这样的东西

if not visited[v]

作为上面显示的主for循环的条件，为了防止递归调用已经访问过的顶点，那么（取决于Python处理递归调用的顺序）顶点2或顶点4将不能对顶点5进行调用，因为5将已经被1或其中的另一个访问过。
如何在允许调用顶点的同时防止循环，只要所述顶点之前没有出现在通向当前顶点的特定路径中？