my solution will not work..it works only if X & Y or on the same side
of subtree..like suppose if path is left node,root,right node..then
the LCS will do..i think first we need to find whether both X & Y r on
the both side of subtree or different sides..depending on tht we need
to find the path and Y..sorry for multiple posts..

On Jan 8, 3:18 am, Newbie <[email protected]> wrote:
> small correction in my post..
>
>
>
>
>
>
>
> > yes..DFS seems a good solution..then look until u get a Z..here is a
> > piece of code..
>
> > void findSum(TreeNode head, Node nodeZ,ArrayList<Integer> buffer) {
>
> > if (head == null || head == nodeZ) return;
>
> > buffer.add(head.data);
>
> > ArrayList<Integer> c1 = (ArrayList<Integer>) buffer.clone();
> > ArrayList<Integer> c2 = (ArrayList<Integer>) buffer.clone();
>
> > findSum(head.left, nodeZ, c1);
> > findSum(head.right, nodeZ, c2);
>
> > }
>
> > then look thu the arraylist to check see if the Y is there..Hope this
> > helps..correct me if I'm wrong..
>
> > On Jan 8, 12:12 am, nishaanth <[email protected]> wrote:
>
> > > How about this solution, Do a DFS on the graph with x as the start node.
>
> > > If you get z , just see if y is in the stack, if its there then it is in 
> > > the
> > > path,else it is not.
>
> > > correct me if i am wrong.
>
> > > On Fri, Jan 7, 2011 at 7:51 PM, juver++ <[email protected]> wrote:
> > > > Heh, problem clearly stated that there a general binary tree, not BST.
>
> > > > --
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>
> > > --
> > > S.Nishaanth,
> > > Computer Science and engineering,
> > > IIT Madras.

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