Count and map a machine tree by messages

A cluster of machines is wired as a tree. Each machine knows only its own id, its parent's id (None for the root) and its children's ids. A machine can talk only to its parent and its children, and only by sending messages that are delivered asynchronously: later, in any order, interleaved with everyone else's messages.

Write the class that runs on each machine:

class Node:
    def __init__(self, node_id, parent, children, network): ...
    def receive_message(self, from_id, message): ...
    def tick(self): ...          # used from level 3; leave it empty for now

The test harness gives you network, which has two methods:

• network.send(self.node_id, to_id, message) queues a message for a neighbour. The message must be a str, and you choose the format. Sending doesn't deliver anything right away. The harness delivers queued messages one at a time, in a random order, by calling receive_message(from_id, message) on the receiver.
• network.report(self.node_id, result) announces a final answer. It's the "print" of this problem.

Task. The harness starts a count by calling receive_message(None, "count") on the root (from_id=None means the message came from outside the cluster). Eventually the root must call network.report(root_id, n) exactly once, where n is the number of machines. No other machine reports.

• Every machine counts itself plus whatever its children report back. A leaf answers straight away.
• A request going down and an answer coming up are different kinds of messages, so encode the difference in your format.
• A count can be started again after one has finished, and it must give the right answer again. Reset or scope your state.

      r            receive_message(None, "count") at r
     / \           r -> a, r -> b      (requests)
    a   b          b -> x              (request)
        |          a -> r "1"; x -> b "1"; b -> r "2"   (answers, in any order)
        x          r reports 4

A node may have tens of thousands of children, so tracking who has answered must be O(1) per answer.