[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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

思路：利用中序遍历，因为正常情况下，中序遍历能得到递增序列，所以找出第一个左边比右边大的节点s1和最后一个右边比左边小的节点s2即可。

/** * 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 = NULL,*s1 = NULL,*s2 = NULL;    void inorder(TreeNode *root){        if(root -> left)            inorder(root -> left);        if(pre != NULL && pre -> val > root -> val){            if(s1 == NULL){                s1 = pre;                s2 = root;            }else                s2 = root;        }        pre = root;        if(root -> right)            inorder(root -> right);    }    void recoverTree(TreeNode *root) {        inorder(root);        int tmp = s1 -> val;        s1 -> val = s2 -> val;        s2 -> val = tmp;    }};