inorder in non binary tree python

Question

我想在 python 中构建一个函数，它获取一些非二叉树并将树中的值从左到右按顺序放在列表中。

Answer 1

Your code isn't including all the child nodes because you're specifically picking out just one node from your first_half and second_half lists to recurse on and ignoring everything else. Probably you want to loop over each of those lists instead:

    mid = int(len(children) / 2) + 1
    first_half = children[:mid]
    second_half = children[mid:]
    for child in first_half:
        child.iters(by_order)
    by_order.append(self)
    for child in second_half
        child.iters(by_order)

It's worth noting that this order seems a bit awkward. You're choosing to stick the parent node in the exact middle of its children (or as close as possible, for odd-numbers of children), but that is a bit arbitrary. While the child nodes have a definitive order, there's not necessarily a clear reason to order the parent in any particular relation to them. Indeed, for some kinds of non-binary trees (like B-trees), there are multiple values stored in each node of the tree, and a proper in-order traversal would interleave them in between the child nodes' values.