在不递归的情况下，在所有到达某个深度的路径中查找 n 叉树的所有唯一节点

Question

I'm trying to write an algorithm in Python to get a unique list of all nodes in a tree where the path reaches a certain depth.

• Each child has an unknown number of children prior to traversal
• The children can be accessed via an iterable (eg for child in B.get_children() )

For example, see this tree (asterisks mark node that should be included):

       A*
       |
     -----
    |     |
    B*    C*
    |     |
    |    ---
    |   |   |
    D*  E   F*
  / | \     | \
 G* H* I*   J* K*
               |
               L

Let's say I'm trying to reach a depth of 3. I need a function that would yield the sequence [G, H, I, J, K, D, F, B, C, A] in any order.

Note the omission of:

• E (doesn't reach depth of 3)
• L (exceeds depth of 3)

I feel there is a way to get this list recursively. Something along the lines of:

def iterate_tree(path: List[T], all_nodes: Set[T]):
    if len(path) == 4:
        for node in path:
            if node not in all_nodes:
                all_nodes.add(node)
                yield node
    else:
        for node in path[-1].get_children():
            path = path.copy()
            path.append(node)
            yield from iterate_tree(path, all_nodes)

iterate_tree([A] ,set())

I don't know if the above works, but I think I can hack it from that. What I don't like about the (probably incorrect) solution is:

1. The recursion: I'm writing this for an unknown depth. I don't want a stack-overflow.
2. I really feel like there must be a way to do this without carrying around a set of previously yielded nodes.
3. I have to make a copy of path at each iteration so I don't mess up other branches of the recursion.

Any suggestions?

Answer 1

For point 1, you can use an explicit stack and loop instead of recursion.

For point 2, I'm not sure I see a problem with keeping a set of yielded nodes. Memory is cheap and if you need to detect duplicates, re-traversing the tree every time you yield is extremely expensive.

Furthermore, your implementation checks for uniqueness based on node hashability, but it's unclear how nodes compute their hash. I assume you're using Node.val for that. If you're hashing based on object reference, "uniqueness" seems pointless since you're guaranteed that a tree of Node objects is unique by identity. The example here doesn't show what a clash on uniqueness would entail. My implementation assumes the hash is object identity (as it should be) and that you can access the value for uniqueness separately using Node.val .

For point 3, if you're working recursively there's no need to copy the path list since you revisit the call frame and can append/pop on a single list. Iteratively, you can keep a parent_of dict alongside the nodes_yielded set that keeps a reference to the parent of each node. When we reach a node at the desired depth, we can walk the links in this dictionary to reconstruct the path, avoiding revisiting a branch more than once thanks to nodes_yielded . A second set, vals_yielded can be used to enforce uniqueness on the yields.

Lastly, I don't really know what your data structures are so in the interest of a minimal, complete example , I've provided something that should be adaptable for you.

import collections

def unique_nodes_on_paths_to_depth(root, depth):
    parent_of = {root: None}
    nodes_yielded = set()
    vals_yielded = set()
    stack = [(root, depth)]

    while stack:
        node, depth = stack.pop()

        if depth == 0:
            while node and node not in nodes_yielded:
                if node.val not in vals_yielded:
                    vals_yielded.add(node.val)
                    yield node

                nodes_yielded.add(node)
                node = parent_of[node]
        elif depth > 0:
            for child in node.children:
                parent_of[child] = node
                stack.append((child, depth - 1))

if __name__ == "__main__":
    """
           A*
           |
         -----
        |     |
        B*    C*
        |     |
        |    ---
        |   |   |
        D*  E   F*
      / | \     | \
     G* H* I*   J* K*
                   |
                   L
    """
    Node = collections.namedtuple("Node", "val children")
    root = Node("A", (
        Node("B", (
            Node("D", (
                Node("G", ()),
                Node("H", ()),
                Node("I", ()),
            ))
        ,)),
        Node("C", (
            Node("E", ()),
            Node("F", (
                Node("J", ()),
                Node("K", (
                    Node("L", ())
                ,)),
            )),
        ))
    ))
    print([x.val for x in unique_nodes_on_paths_to_depth(root, 3)])
    #      => ['K', 'F', 'C', 'A', 'J', 'I', 'D', 'B', 'H', 'G']