What I can think
is case is :
1
/ \
2 3
/ \
4 5
/ \ \
6 7 8
/ \ \
9 a b
so from a->b is
a->7->4->2->5->8->b
On Sat, Jun 23, 2012 at 2:44 PM, Avinash <[email protected]> wrote:
> Some how I found that we need to run bfs twice to get the largest distance
> between any two nodes of a tree. Please explain me how it works.
> regards,
> avinash
>
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Regards
Kumar Vishal
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