题目：

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：

中序遍历整个二叉搜索树，若发现数值未按序排列，则为有问题的节点。最后还得考虑只有两个节点的情况。

代码：

/**
 * 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* res1;
    TreeNode* res2;
    TreeNode* res3;
    TreeNode* pre;
    TreeNode* cur;
    int num;
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        num = INT_MIN;
        res1 = NULL;
        res2 = NULL;
        res3 = NULL;
        pre = NULL;
        cur = NULL;
        searchTree(root);
        if(res2 == NULL)
        {
            int t = res1->val;
            res1->val = res3->val;
            res3->val = t;
        }
        else
        {
            int t = res1->val;
            res1->val = res2->val;
            res2->val = t;
        }
    }
    
    void searchTree(TreeNode* root)
    {
        if(root == NULL)
        {
            return;
        }
        else
        {
            searchTree(root->left);
            pre = cur;
            cur = root;
            if(res1 == NULL)
            {
                if(pre != NULL && pre->val > cur->val)
                {
                    res1 = pre;
                    res3 = cur;
                }
            }
            else if(res2 == NULL)
            {
                if(pre != NULL && pre->val > cur->val)
                {
                    res2 = cur;
                }
            }
            searchTree(root->right);
        }
    }
};