OK

On Sun, Jul 17, 2011 at 2:59 PM, saurabh singh <[email protected]> wrote:

> 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 <[email protected]>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 <[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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>
>
>
> --
> Saurabh Singh
> B.Tech (Computer Science)
> MNNIT ALLAHABAD
>
>
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>



-- 
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD

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