题解 | #二叉搜索树与双向链表#

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
	void InOrderConvert(TreeNode*cur,TreeNode*&prev)
	{
		if(cur==nullptr)
		{
			return;
		}
		InOrderConvert(cur->left, prev);
		cur->left=prev; 
		if(prev)
		{
			prev->right=cur;
		}
		prev=cur;
		InOrderConvert(cur->right, prev);
	} 
    TreeNode* Convert(TreeNode* pRootOfTree) {
       //这个题有很多限制 只能动指针了
	   if(pRootOfTree==nullptr)
	   {
		return nullptr;
	   }
	   TreeNode*prev=nullptr;
	   TreeNode*cur=pRootOfTree;
		InOrderConvert(cur, prev);
		while(cur->left)
		{
			cur=cur->left;
		}
		return cur;
    }
};