Ok lets see.
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 :(


On Mon, Jul 11, 2011 at 3:16 PM, aanchal goyal <goyal.aanch...@gmail.com>wrote:

> we should not deform the tree.
> - converting into dll and solving.
> - doing inorder and hashing
> - doing inorder and saving in array
> All above solutions I know, so dont post them,
> i dont know how to solve this using inorder and reverse inorder approach..
>
>
> On Mon, Jul 11, 2011 at 3:13 PM, Piyush Sinha <ecstasy.piy...@gmail.com>wrote:
>
>>
>> If we dont want the tree back, we can convert the BST to DLL and do the
>> job..
>> On Mon, Jul 11, 2011 at 3:01 PM, aanchal goyal 
>> <goyal.aanch...@gmail.com>wrote:
>>
>>> Given a BST and integer value K. Find all pairs of nodes (x,y), such that
>>> x->data + y->data = K
>>> Time O(n)
>>>
>>> Can someone provide a pseudocode/code to solve this using the concept of
>>>  inorder and reverse inorder traversal of BST?
>>> PS: please don't post other solutions for this, I know this can be solved
>>> in other ways too. I am not able to code this using the above concept..
>>> --
>>> Regards,*
>>> Aanchal Goyal*.
>>>
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>>
>>
>>
>> --
>> *Piyush Sinha*
>> *IIIT, Allahabad*
>> *+91-8792136657*
>> *+91-7483122727*
>> *https://www.facebook.com/profile.php?id=100000655377926 *
>>
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>
>
>
> --
> Regards,*
> Aanchal Goyal*.
>
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-- 
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD

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