@Dave
i think your solution won't work
consider inorder traversal of a BST is 1 6 7 8 15 and x = 14
initially both u,v (1,1)
according to u your algorithm will proceed like
(1,1) -> (1,6) -> (1,7) -> (1,8) -> (1,15) -> (6,15) ............ -> (15,15)

and clearly in second step of your solution if (u+v) > x after advancing u
still u+v will be greater than x
so something is wrong
I think your solution will work in case we need to find 2 nodes with
difference x.

correct me if i am wrong.!!

On Mon, Jun 27, 2011 at 6:13 PM, Dave <[email protected]> wrote:

> @Nishant: No need to store the data in an array. Do two inorder
> traversals simultaneously. Let u and v be the current nodes of the two
> traversals, respectively. If u + v < x, then advance the "v"
> traversal. If u + v > x, advance the "u" traversal.
>
> Dave
>
> On Jun 27, 3:40 am, Nishant Mittal <[email protected]> wrote:
> > do inorder traversal of tree and store values in an array.
> > Now find pairs by applying binary search on array..
> >
> > On 6/27/11, manish kapur <[email protected]> wrote:
> >
> >
> >
> >
> >
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-- 
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee

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