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 _Convert(TreeNode* cur,TreeNode*& prev)//必须加引用才能保证只有一个prev
	{
		if(!cur)return ;
		_Convert(cur->left,prev);
		///
		cur->left=prev;
		if(prev)
		prev->right=cur;
		prev=cur;
		_Convert(cur->right,prev);
	}
    TreeNode* Convert(TreeNode* root) {
		if(!root)return nullptr;
        TreeNode* prev=nullptr;
		_Convert(root,prev);
		TreeNode* ans=root;
		while(ans->left)
		ans=ans->left;
		return ans;
    }
};