Dave,

I am unable to write code for this so i am asking your help.

Thanks,
Swathi

On Mon, Jun 27, 2011 at 11:28 PM, Dave <[email protected]> wrote:

> @Swathi: No. I think the high level description should be adequate for
> you to write your own code or pseudocode, albeit recognizing that you
> may have to look up how to do an inorder traversal using a stack
> instead of recursion.
>
> Dave
>
> On Jun 27, 9:34 am, Swathi <[email protected]> wrote:
> > Dave,
> >
> > Can you provide the psuedo code for this..
> >
> > Thanks,
> > Swathi
> >
> >
> >
> > On Mon, Jun 27, 2011 at 7:30 PM, Dave <[email protected]> wrote:
> > > @Sunny. Mea culpa. You are correct. Revised (and correct) algorithm.
> > > Do two inorder traversals, one in the usual (descend to the left
> > > before descendung to the right) direction and the other in the
> > > reversed(descend to the right before descending to the left)
> > > direction. Let u and r be the current nodee of the two traversals,
> > > respectively. If u + r < x, then advance the usual traversal and
> > > repeat the comparison. If u + r > x, advance the reverse traversal and
> > > repeat the comparison. If u + r = x, and if u != r, then terminate
> > > with success. If u = r, then terminate with failure.
> >
> > > Dave
> >
> > > On Jun 27, 7:53 am, sunny agrawal <[email protected]> wrote:
> > > > @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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