如何找到树的任意两个顶点之间的边或顶点数量？

Question

Its a general tree (acyclic graph), so there can be only one such path. What algorithm can I use for this?

EDIT:

Need to find the paths for all pairs of vertices in the tree

Answer 1

I want to extend @templatetypedef's answer here ¹ .

Note that in your problem, you need to do at least one write per each pair of nodes in the tree. There are n*(n-1)/2 of these.
Thus, in terms of big O notation, you cannot find an algorithm that runs better than O(n^2) .

Now, use DFS or BFS to find the path per node. It will run in O(n+m) ( n vertices, m edges). But since it is a tree, we know that m=n-1 , giving us O(n) for BFS/DFS. Note that in a single BFS/DFS from some node v - you get d(v,u) for EVERY node u .

If you repeat it per each node, it will get you O(n^2) - which is optimal in terms of big O notation. I do agree you might get better constants with some optimizations, but that's about it.

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(1) Started it as a comment, but it got too long and I figured it worth an answer.

Answer 2

One simple option is to do a depth-first search starting at one of the nodes until you reach the other node. You can then look at the path that was taken from the first node to the second, which must be the unique path between those nodes, and then count how many edges are in that path. This runs in linear time because a DFS on a tree only takes linear time.

Hope this helps!

Answer 3

You need to find the Lowest Common Ancestor (LCA). There are different approaches as you can study here: http://leetcode.com/2011/07/lowest-common-ancestor-of-a-binary-tree-part-i.html

I have used the fallowing solution since I'm not fan to recursivity, that should work better for your problem since you need to find all the pairs in an efficient manner.

1-) Finding path between A -> B : -Iterate starting from node A going up each parent node flagging each one with A Flag, until parent root fund. -Iterate starting from node B going up until A Flag found. You have found LCA node. - Result path is a list from A to LCA node, plus a reversed list from B to LCA node.

2-) Improvement finding A -> B : - Iterate both nodes simultaneously, flagging with A Flag, and B Flag each ancestor node. Until A iterations find B Flag or B iterations find A Flag. Then the first to find another node flag has found the LCA node.

3-) Finding all pairs paths: You can simply use solution above for each pair. Or try to consider making a first pass iterating all the tree creating lists of flags, then a second pass to identify all LCA nodes for each pair, that would be inconvenient when the tree number of nodes grows too big.