[leetcode][BST] Recover Binary Search Tree

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? //还没想到

分析：本质中序遍历序列是排序的。

当前值比前一个值小时，若还没找到第一个数，则前一个值就是第一个发生逆序的值；若已经找到第二个数，则当前值就是第二个发生逆序的值。

/**
 * 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 (NULL == root || NULL == root->left && NULL == root->right) return;
	stack<TreeNode *> sta;
	TreeNode *p = root;
	bool foundFirst = false;//是否已找到第一个逆序元素
	TreeNode *lastNode = NULL;
	TreeNode *nextNode = NULL;//用于处理当中序遍历序列相邻两个元素交换的情况
	TreeNode *p1 = NULL, *p2 = NULL;
	while (p != NULL || !sta.empty()){
		while (p != NULL){
			sta.push(p);
			p = p->left;
		}
		TreeNode *q = sta.top();
		sta.pop();
		if (lastNode != NULL && q->val < lastNode->val){
			if (!foundFirst){
				p1 = lastNode;
				nextNode = q;
				foundFirst = true;
			}
			else if(lastNode != p1){
				p2 = q;
				break;
			}
		}
		lastNode = q;
		p = q->right;
	}
	if (NULL == p2) p2 = nextNode;
	int tmp = p1->val;
	p1->val = p2->val;
	p2->val = tmp;
}
};