LeetCode-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}".

对二叉搜索树中序遍历的话应该得到一个升序序列，这个序列中两个逆序的地方就是所要找的地方。

class Solution {
public:
    void check(TreeNode* root,TreeNode** pre,TreeNode** n1,TreeNode** n2){
        if(root==NULL)return ;
        check(root->left,pre,n1,n2);
            if((*pre)!=NULL&&(*pre)->val>root->val){
                if((*n1)==NULL){
                    *n1=*pre;
                }
                *n2=root;
            }
			(*pre)=root;
        check(root->right,pre,n1,n2);    
    }
    void recoverTree(TreeNode *root) 
    {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        TreeNode * pre=NULL,*n1=NULL,*n2=NULL;
		check(root,&pre,&n1,&n2);
		int i=n1->val;
        n1->val=n2->val;
        n2->val=i;
    }
};