Proof of correctness: Algorithm for diameter of a tree in graph theory

Question

In order to find the diameter of a tree I can take any node from the tree, perform BFS to find a node which is farthest away from it and then perform BFS on that node. The greatest distance from the second BFS will yield the diameter.

I am not sure how to prove this, though? I have tried using induction on the number of nodes, but there are too many cases.

Any ideas would be much appreciated...

Answer 1

Let's call the endpoint found by the first BFS x. The crucial step is proving that the x found in this first step always "works" -- that is, that it is always at one end of some longest path. (Note that in general there can be more than one equally-longest path.) If we can establish this, it's straightforward to see that a BFS rooted at x will find some node as far as possible from x, which must therefore be an overall longest path.

Hint: Suppose (to the contrary) that there is a longer path between two vertices u and v, neither of which is x.

Observe that, on the unique path between u and v, there must be some highest (closest to the root) vertex h. There are two possibilities: either h is on the path from the root of the BFS to x, or it is not. Show a contradiction by showing that in both cases, the u-v path can be made at least as long by replacing some path segment in it with a path to x.

[EDIT] Actually, it may not be necessary to treat the 2 cases separately after all. But I often find it easier to break a configuration into several (or even many) cases, and treat each one separately. Here, the case where h is on the path from the BFS root to x is easier to handle, and gives a clue for the other case.

[EDIT 2] Coming back to this later, it now seems to me that the two cases that need to be considered are (i) the u-v path intersects the path from the root to x (at some vertex y, not necessarily at the u-v path's highest point h); and (ii) it doesn't. We still need h to prove each case.