OK...
suppose our tree is
5
/ \
4 6
/ \
3 7
/ \
2 8
/ \
1 9
now k=10;
so will it return all the pairs like 1,9 2,8 . . ..5,5. . . .. .8,2 9,1 ??
On Sun, Jul 17, 2011 at 7:00 AM, saurabh singh <[email protected]> wrote:
> @sagar This is what Dave is suggesting in a more pseudocode way:-
>
> 1->Traverse a pointer right down to the leftmost element,i.e.the
> shortest,say small
> 2->traverse a pointer left down to the rightmost element i.e.the
> largest.say
> large
> while(small!=large)
> 3->Compare their sum.If sum>k set large to its successor in reverse
> inorder.(I am not sure if u meant the same but I am assuming rev inorder
> to
> be right->node->left)
> else set small to its inorder successor.
> break when u get the desired k.
> print :)
> return
> if u get out of the loop without getting the number
> then such number does not exist.print :(
>
> Amortized complexity order n.
>
>
>
> --
> Saurabh Singh
> B.Tech (Computer Science)
> 5h sem
> MNNIT ALLAHABAD
>
>
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**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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