I have a question which is part of my program.
For a tree T=(V,E) we need to find the node v in the tree that minimize the length of the longest path from v to any o
There are two approaches to do this (both runs in the same time):
Select any vertex v1
on your tree. Run BFS from this vertex. The last vertex (v2
) you will proceed will be the furthest vertex from v1
. Now run another BFS, this time from vertex v2
and get the last vertex v3
.
The path from v2
to v3
is the diameter of the tree and your center lies somewhere on it. More precisely it lies in the middle of it. If the path has 2n + 1
points, there will be only 1 center (in the position n + 1
). If the path has 2n
points, there will be two centers at the positions n
and n + 1
.
You only use 2 BFS calls which runs in 2 * O(V)
time.