99. Recover Binary Search Tree（难）

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?

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有两种情况，如果是中序遍历相邻的两个元素被调换了，

很容易想到就只需会出现一次违反情况，只需要把这个两个节点记录下来最后调换值就可以；

如果是不相邻的两个元素被调换了，举个例子很容易可以发现，会发生两次逆序的情况，那么这时候需要调换的元素应该是第一次逆序前面的元素，和第二次逆序后面的元素。比如1234567,1和5调换了，会得到5234167，逆序发生在52和41，我们需要把4和1调过来，那么就是52的第一个元素，41的第二个元素调换即可。

/**
 * 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) {
		if (root == NULL) return;
		TreeNode* pre = NULL;
		vector res;
		dfs(root, pre, res);
		if (!res.empty()){
			swap(res[0]->val, res[1]->val);
		}
	}
	void dfs(TreeNode* root, TreeNode* &pre, vector &res){
		if (root == NULL) return;
		dfs(root->left, pre, res);
		if (pre != NULL&&pre->val > root->val){
			if (res.empty()){
				res.push_back(pre);
				res.push_back(root);
			}
			else{
				res[1] = root;
			}
		}
		pre = root;
		dfs(root->right, pre, res);
	}
};