蟒蛇最低的共同祖先 [英] python lowest common ancestor

问题描述

什么是在Python实现最低的共同祖先的最简单的方法？我有重新由具有指向其父每个节点psented $ P $一棵树，我希望能够找到给两个节点的共同祖先。我已经想出了一些想法，但没有一个是特别有吸引力

1. 让每个节点包含其基地名单，并进行联接，找到最长的共同preFIX然后取最后一个元素。不幸的是，我不知道任何内置的方式做到最长的公共preFIX所以这需要手动循环。

2. 让每个节点包含了一组其基地和执行交集，并采取最大元素。但是，这需要定义自定义比较运算，我甚至不知道是否会工作。

我应该怎么办？我在找的东西，有利于简便了性能，因此需要复杂的处理解决方案都出来了。

编辑：我发现，虽然没有一个内置的方式，你可以用拉链做最长的共同preFIX在一行中，所以它仍然是相当简单

。

 普通= [X对于x的ZIP（* baselists）如果len（集（X））== 1] [ -  1]
 

解决方案

根据的假设，你不能修改你的树，包括深度，你可以做到以下几点：

对于每个节点，递归遍历树向上，直到你打根。在每个父节点，接入节点到列表。这应该给你 list_a 和 list_b 。遍历最短的列表，比较从每个列表元素。当你发现一个不匹配，previous项即是最大的父元素。

What is the easiest way to implement lowest common ancestor in Python? I have a tree which is represented by each node having a pointer to its parent, and I want to be able to find the first common ancestor given two nodes. I've come up with several ideas but none are particularly appealing

1. Have each node contain a list of its bases, and to perform a join, find the longest common prefix and then take the last element. Unfortunately, I don't know of any builtin way to do longest common prefix so this requires manual looping.

2. Have each node contain a set of its bases and perform a set intersection, and take the maximum element. But this requires defining custom comparison operators, and I'm not even sure if it would work.

What should I do? I'm looking for something that favors simplicity over performance, so solutions requiring complicated processing are out.

Edit: I discovered that while there isn't a builtin way, you can do longest common prefix in one line using zip, so it's still fairly simple.

common = [x for x in zip(*baselists) if len(set(x)) == 1][-1]

解决方案

Under the assumption that you cannot amend your tree to include depth, you can do the following:

For each Node, recursively traverse the tree upwards until you hit the root. At each parent node, insert the node into a list. This should give you list_a and list_b. Iterate over the shortest list, comparing elements from each list. When you find one that doesn't match, the previous entry is your largest parent element.

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