Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.
Tell us the 2 values swapping which the tree will be restored.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?
 
Example :

Input : 
         1
        / \
       2   3

Output : 
       [1, 2]

Explanation : Swapping 1 and 2 will change the BST to be 
         2
        / \
       1   3
which is a valid BST         
Solution:

Time: O(n)
Space: O(h)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) {
 *      val = x;
 *      left=null;
 *      right=null;
 *     }
 * }
 */
public class Solution {
    public ArrayList<Integer> recoverTree(TreeNode A) {
        Integer prev = null;
        Integer one = null;
        Integer two = null;
        ArrayList<Integer> result = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode curr = A;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            if (!stack.isEmpty()) {
                curr = stack.pop();
                int val = curr.val;
                if (prev != null) {
                    if (prev > val) {
                        if (one == null) {
                            one = prev;
                        }
                        two = val;
                    }
                }
                prev = val;
                curr = curr.right;
            }
        }
        result.add(two);
        result.add(one);
        return result;
    }
}