LeetCode Recover Binary Search Tree

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p
	p = 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;
	}
}