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 <[email protected]> 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 <
> [email protected]> 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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