yes..DFS seems a good solution..then look until u get a Z..here is a
piece of code..
static ArrayList<Integer> buffer;
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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