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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