面试题20：搜索二叉树可能有两个元素发生了交换，如何恢复BST？

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

 /**
  * 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* mistake1 = nullptr;
     TreeNode* mistake2 = nullptr;
     void recoverTree(TreeNode *root) {
         inorder(root);
         if(mistake1 && mistake2) swap(mistake1->val,mistake2->val);
     }
     void inorder(TreeNode* root){
         if(root == nullptr) return ;
         inorder(root->left);
         if(pre == nullptr) pre = root;
         else{
             if(pre->val > root->val){
                 if(mistake1 == nullptr) mistake1=pre;
                 mistake2 = root;
             }
             pre = root;
         }
         inorder(root->right);
     }
 };