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:
    TreeNode* pre = nullptr;
    TreeNode* p = nullptr;
    TreeNode* q =nullptr;
    void recoverTree(TreeNode *root) {
        helper(root);
        swap(p->val,q->val);
    }
     
    void helper(TreeNode* root) {
        if(!root) return;
        helper(root->left);
        if(pre && pre->val > root->val){
            if(!p)
            p = pre;
            q = root;
        }
        pre = root;
        helper(root->right);
    }
};