Chonku, you can do that only when you have the links to parent nodes. I
couldn't come up with a way of doing what you said on a basic BST(nodes
having pointers only to their 2 children) that is why I suggested using an
array. It doesn't change the overall complexity but if you have an idea
about how implement your idea on a basic BST, I would like to hear it.
Thanks,
Seckin
On Fri, Aug 6, 2010 at 2:56 AM, sharad kumar <aryansmit3...@gmail.com>wrote:
> do the inorder traversal of the bst ...this gives the sorted array......
> from that use
>
> int i=0,j=length(array)
> while(i<j)
> {
> if(array[i]+array[j]>sum)
> --j;
> else if(array[i]+array[j]<sum)
> ++i;
> else if((array[i]+array[j])==sum)
> return i,j
> else
> ++i,--j;
> }
>
>
> On Fri, Aug 6, 2010 at 3:10 PM, Chonku <cho...@gmail.com> wrote:
>
>> Two inorders would achieve the same thing without using an array. One
>> pointer running inorder with LDR and other pointer running inorder with RDL.
>> Compare the sum at the two nodes and then adjust them accordingly.
>>
>> On Fri, Aug 6, 2010 at 2:11 PM, Manjunath Manohar <
>> manjunath.n...@gmail.com> wrote:
>>
>>> the solution elegant..but is there any on the fly method by just
>>> exploiting the BST property....by using left and right pointers
>>>
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