9 1 is analogous to 1 9...And the question requires only two nodes,it does
not says about all such pairs.

On Sun, Jul 17, 2011 at 2:52 PM, sagar pareek <sagarpar...@gmail.com> wrote:

> 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 <saurab...@gmail.com>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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-- 
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD

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