[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?

利用中序周游从小到大的顺序检查，找到第一个异常元素。再利用中序周游从大到小的顺序检查，找到第一个异常元素。

将这两个元素交换即可。

class Solution {
public:
    TreeNode *tmp;
    bool tag;
    bool key;
    TreeNode *tmp1;
    TreeNode *tmp2;
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        tmp1 = NULL;
        tmp2 = NULL;
        tag = false;
        key = true;
        check(root);
        tag = false;
        key = true;
        check1(root);
        if(tmp1 != NULL && tmp2 != NULL)
        {
            int m = tmp1->val;
            tmp1->val = tmp2->val;
            tmp2->val = m;
        }
    }
    void check(TreeNode *root)
    {
        if(key == false)
            return;
        if(root == NULL)
            return;
        check(root->left);
        if(tag == false)
        {
            tmp = root;
            tag = true;
        }
        else
        {
            if(root->val <= tmp->val)
            {
                tmp1 = tmp;
                key = false;
            }
            else
                tmp = root;
        }
        check(root->right);
    }
    void check1(TreeNode *root)
    {
        if(key == false)
            return;
        if(root == NULL)
            return;
        check1(root->right);
        if(tag == false)
        {
            tmp = root;
            tag = true;
        }
        else
        {
            if(root->val >= tmp->val)
            {
                tmp2 = tmp;
                key = false;
            }
            else
                tmp = root;
        }
        check1(root->left);
    }
};