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) {

		TreeNode *prev=nullptr;
		InOrderConvert(pRootOfTree, prev);
		TreeNode*head=pRootOfTree;
		while(head&&head->left)
		head=head->left;
		return head;
        
    }
};