[leetcode]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?

这题想了比较久=.=
第一感觉是用map<TreeNode*,int> 存下中序遍历的结果。再交换错误的两个。但是不太符合题意。

比如说一个BST用中序遍历的结果是:
1,2,3,4,5,6,7
其中有两个搞混了(2和4搞混了)：
1,4,3,2,5,6,7
当检查到3时就会发现出现了错误(应该是严格单调增的)，将3和他前面的4记录下来：{4,3}
在当检查到2时又出现了错误，第二次发现错误将结果改为{4,2}
至多出现两次错误，所以答案就是{4,2}
但是也有可能是相邻两个数混了(5和6混了):
1,2,3,4,6,5,7
这样只会出现一次错误，结果就是{6,5}
所以还需要记录在某个节点之前的节点before

/** * 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 *a = NULL;    TreeNode *b = NULL;    TreeNode *before = NULL;    void find(TreeNode *root){        if(root == NULL) return;        if(root->left!=NULL) find(root->left);        if(before!=NULL&&root->val<before->val) {            if(a==NULL){                a = before;                b = root;            }            else{                b = root;            }        }        before = root;        if(root->right!=NULL) find(root->right);    }    void recoverTree(TreeNode *root) {        //if(root == NULL) return;        int tem;        find(root);        if(a!=NULL){            tem = b->val;            b->val = a->val;            a->val = tem;        }    }};