i think it can done like this,
assume we have to find 'x' and 'y'  wer s='x'+'y'

1) select ith node from tree (from beginning to end)
2) y= s - " ith number (node)"
3) now search for 'y' in BST if we found then there is node such that s= x +
y; otherwise no..

-- Prashant Kulkarni
|| Lokaha Samastaha Sukhino Bhavanthu ||
|| Sarve Jana Sukhino Bhavanthu ||



On Fri, May 14, 2010 at 2:47 AM, divya jain <sweetdivya....@gmail.com>wrote:

> form a sorted  array from inorder traversal of tree. now take to pointers
> one to the beginning of array and other at the end. now check if the sum of
> element is greater than reqd sum then increment 1st ptr. if their sum is
> less than reqd sum then decrement 2nd ptr. if their sum is equal to the reqd
> sum then this is the ans..
> hope it will work..
>
>
> On 13 May 2010 20:11, jalaj jaiswal <jalaj.jaiswa...@gmail.com> wrote:
>
>> given a bst... find two nodes whose sum is equal to a number k ... in O(n)
>> time and constant space...
>>
>> --
>> With Regards,
>> Jalaj Jaiswal
>> +919026283397
>> B.TECH IT
>> IIIT ALLAHABAD
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