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);
    }
};