Recover Binary Search Tree - LeetCode

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Discuss

重点在中序遍历，注意这一步 

if (prev != null && prev.val > root.val) {
                if (first == null) {
                    first = prev;
                }

                second = root;
            }

以下是AC代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {

        private TreeNode first;
        private TreeNode second;
        private TreeNode prev;

        public void recoverTree(TreeNode root) {
            if (root == null) {
                return;
            }

            first = null;
            second = null;
            prev = null;

            recoverTreeHelper(root);
            switchValue(first, second);
        }

        private void switchValue(TreeNode node1, TreeNode node2) {
            int val = node1.val;
            node1.val = node2.val;
            node2.val = val;
        }

        private void recoverTreeHelper(TreeNode root) {
            if (root == null) {
                return;
            }

            recoverTreeHelper(root.left);

            if (prev != null && prev.val > root.val) {
                if (first == null) {
                    first = prev;
                }

                second = root;
            }

            prev = root;
            recoverTreeHelper(root.right);
        }

}