Leetcode 99. Recover Binary Search Tree O(1)

题目中要求用constant space去结题，那就不能使用中序遍历类似的递归写法，因为这些的空间复杂度平均水平是O(logN)。那么只有使用一种（新的）遍历算法Morris Traversal。

然后结合中序遍历的结题思路，左子树的最大值要小于根节点和右子树的值。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    void recoverTree(TreeNode* root) {        TreeNode* cur = root,* pre = nullptr;        while(cur)        {            if(cur->left == nullptr)            {                iserror(cur);                last = cur;                //printf("%d\n", last->val);                cur = cur->right;            }            else            {                pre = cur->left;                while(pre->right != nullptr && pre->right != cur)                    pre = pre->right;                if(pre->right == nullptr)                {                    pre->right = cur;                    cur = cur->left;                }                else                {                    pre->right = nullptr;                    iserror(cur);                    last = cur;                    //printf("%d\n", last->val);                    cur = cur->right;                }            }        }        swap(node1->val, node2->val);    }private:    TreeNode* node1 = nullptr;    TreeNode* node2 = nullptr;    TreeNode* last = new TreeNode(INT_MIN);    void iserror(TreeNode* cur)    {        //printf("%d %d\n", last->val, cur->val);        if(!node1 && last->val > cur->val)   node1 = last;        if( node1 && last->val > cur->val)   node2 = cur;    }};