What I can think
is case is :
10
/ \
6 13
/ \
4 5
/ \ \
6 7 8
/ \ \
9 a b
so from a->b is
a->7->4->2->5->8->b
1-> Left Tree then
2-> Right Tree
add them
On Sat, Jun 23, 2012 at 3:49 PM, Kumar Vishal <kumar...@gmail.com> wrote:
> 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 <jain.av...@gmail.com> 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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