Increasing Order Search Tree

1/**
2 * Definition for a binary tree node.
3 * public class TreeNode {
4 *     int val;
5 *     TreeNode left;
6 *     TreeNode right;
7 *     TreeNode() {}
8 *     TreeNode(int val) { this.val = val; }
9 *     TreeNode(int val, TreeNode left, TreeNode right) {
10 *         this.val = val;
11 *         this.left = left;
12 *         this.right = right;
13 *     }
14 * }
15 */
16class Solution {
17    public TreeNode increasingBST(TreeNode root) {
18        TreeNode curr = root;
19        Stack<TreeNode> stack = new Stack<TreeNode>();
20        List<TreeNode> list = new LinkedList<TreeNode>();
21
22        while (curr != null || !stack.isEmpty()) {
23            if (curr != null) {
24                stack.push(curr);
25                curr = curr.left;
26            }
27            else {
28                curr = stack.pop();
29                list.add(curr);
30                curr = curr.right;
31            }
32        }
33
34        for (int i = 0; i < list.size(); i++) {
35            TreeNode currentNode = list.get(i);
36
37            currentNode.left = null;
38            // the following conditional is actually not even necessary if
39            // your loop condition is < list.size() - 1, since you know for
40            // sure the rightmost Node, with the highest value, will not
41            // have a right child. 
42            if (i < list.size() - 1) {
43                currentNode.right = list.get(i + 1);
44            }
45            else {
46                currentNode.right = null;
47            }
48        }
49
50        return list.get(0);
51    }
52}