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}"

.
修正二叉树中错误的两个结点，在不改变结构的情况下。
思路：首先想到的就是中序遍历二叉搜索树，中序遍历二叉搜索树的结点是按从小到大的顺序，如果出现该节点值比前一个结点小，则说明这个值不合法。用pPre存中序遍历的前一个结点，方便比较大小，first和second分别保存较大的和较小的值。

/**
* 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 recoverMistake(TreeNode *root,TreeNode *&first,TreeNode *&second,TreeNode *&pPre)
{
if(root==NULL)
return;
recoverMistake(root->left,first,second,pPre);
if(pPre && pPre->val>root->val)
{
if(first==NULL)
{
first=pPre;
second=root;
}
else
{
second=root;
}
}
pPre=root;
recoverMistake(root->right,first,second,pPre);
}
void recoverTree(TreeNode *root) {
TreeNode *first=NULL,*second=NULL,*pPre=NULL;
recoverMistake(root,first,second,pPre);
swap(first->val,second->val);
}
};