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?

BST的题目一般都是和中序遍历有关，因为他中序遍历得到的一定是有序序列。

此题就是中序遍历的过程中记录出现非递增的情况，记录下非递增的2个节点，然后交换它们的数据。

//中序遍历，p1，p2分别记录出错的两个节点//p记录中序遍历时的前一个节点TreeNode* p1=NULL, *p2=NULL, *p=NULL;void recover(TreeNode* root){if(NULL == root)return ;if(root->left != NULL)recover(root->left);if(p!=NULL && root->val<p->val){if(p1 == NULL){//p1，p2是相邻节点p1 = p;p2 = root;}else{//p1和p2不相邻p2 = root;}}//中序遍历的节点记录到pp = root;if(root->right != NULL)recover(root->right);}void recoverTree(TreeNode *root) {if(NULL == root){return ;}recover(root);if(p1 && p2){int v = p1->val;p1->val = p2->val;p2->val = v;}}