leetcode---recover-binary-search-tree---树

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:    TreeNode *p1 = NULL, *p2 = NULL, *pre = NULL;    void dfs(TreeNode *root)    {        if(!root)            return;        dfs(root->left);        if(pre && root->val < pre->val)        {            if(!p1)            {                p1 = pre;                p2 = root;            }            else                p2 = root;        }        pre = root;        dfs(root->right);    }    void recoverTree(TreeNode *root)     {        dfs(root);        if(p1 && p2)            swap(p1->val, p2->val);    }};