问题
I'm curious if there is a quick way to find if an edge exists which is not a part of any possible diameter of a n-ary tree. For example in the following tree, A-B edge will not be a part of any diameter.
I tried by listing down all the possible diameters, but that takes a lot of time and I'm certain that there is a faster way.
回答1:
Let's begin with a simpler question: how would we find any diameter of the tree? One way to do this would be to pick some node and to root the tree at that node. A diameter of the graph then could be found in one of two ways:
- The diameter might purely be contained within one of those subtrees.
- The diameter might be found by taking the roots of two subtrees, computing the longest path starting at each of the subtree roots, and then joining them together through the overall tree root.
So imagine that we recursively visit each subtree and obtain, from each, both the longest path starting at the root of that subtree (we could store this implicitly by having each node store its height) and the length of the longest path purely within that subtree (possibly stored implicitly by tagging each subtree with the length of the longest path within that tree). Once we have this information, we can find some diameter as follows: the diameter is either
- the longest path purely within one of those subtrees, or
- formed by joining two of the longest paths starting at the roots of the subtrees through the root.
All this information can be computed in time O(n) with O(n) auxiliary storage, and so if we just need to determine what the diameter is, we can do so fairly quickly.
Now, let's modify this to actually find all the edges that might get used. We can do this by starting at the root node. Consider the lengths of the paths obtained via routes (1) and (2). If route (1) produces a strictly longer path than route (2), we can recursively descend into each subtree containing a path of that length and run the same process to identify the edges that could potentially be used. If route (2) produces a strictly longer path than route (2), we'd then mark the edge from the root to each of its children who have the longest path starting at a subtree root as being used, and if there's exactly one such subtree we'd then mark each subtree tied for the second-longest path as being used. We'd then recursively descend into those subtrees, always taking paths down subtrees containing one of the many possible longest paths.
This second propagation step takes time O(n) because each node is visited exactly once and the work done is proportional to the number of children. Overall, this is an O(n)-time algorithm that uses O(n) space.
来源：https://stackoverflow.com/questions/54045279/find-the-edge-which-is-not-a-part-of-any-possible-diameter-of-a-tree