返回 n 叉树中从根到叶的所有路径

Question

There's a lot out there about how to print all paths from root to leaf in a tree, but I'm trying to return them. Each node has a list of children, and I need to return a list of strings where each string represents a path. Ie with the tree

   A
  /\
  B F
/ | \
C D  E 

I should return ['A, B, C', 'A,B,D', 'A,B,E', 'A,F'] The function must be recursive and done in python. I've tried this function for a while, but am getting stuck wrapping my head around the recursion. Ie if my function starts with path="", then every time I recall the function I lose that path...

*Edit: solved with help from @flakes's answer and modifying yield [root.value] + path to yield [root.value+", "+path[0])

Answer 1

Start by defining a class for your node structure. It should have a value, and a list of child nodes:

class Node:
    def __init__(self, value, children=None):
        self.value = value
        self.children = children or []

For returning data from recursive functions, I like to use generators as they can simplify the logic a lot.

The first use case is when there are no children. The result should be [value] .

When there are children, you want to loop each child in the children list, find their respective paths, and for each path, append the current [value] to their result.

You might get something like this:

def find_paths(root):
    if len(root.children) == 0:
        yield [root.value]

    for child in root.children:
        for path in find_paths(child):
            yield [root.value] + path

Test case:

tree = Node('a', [Node('b', [Node('c'), Node('d'), Node('e')]), Node('f')])

for path in find_paths(tree):
    print(path)

['a', 'b', 'c']
['a', 'b', 'd']
['a', 'b', 'e']
['a', 'f']

---

If you need it as one flat list, then I would just flatten the result after the call:

output = [value for path in find_paths(tree) for value in path]