99 Recover Binary Search Tree

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode *root) {
        TreeNode *pre = NULL;
        TreeNode *first = NULL;
        TreeNode *second = NULL;
        inorderTraversal(root,pre,first,second);
        int tmp = first->val;
        first->val = second->val;
        second->val = tmp;
    }
    void inorderTraversal(TreeNode *root, TreeNode* & pre, TreeNode *&first, TreeNode* &second)
    {
        if (!root)
            return;
        inorderTraversal(root->left,pre,first,second);
        if (pre && pre->val>root->val)
        {
            if (!first)
                first = pre;
            second = root;
        }
        pre = root;
        inorderTraversal(root->right,pre,first,second);
    }
};