BST转换成有序链表

把二元查找树转变成排序的双向链表(树)
题目:
输入一棵二元查找树,将该二元查找树转换成一个排序的双向链表。
要求不能创建任何新的结点,只调整指针的指向。

 struct BSTreeNode{
     int val;
     BSTreeNode* left;
     BSTreeNode* right;
     BSTreeNode(int x) : val(x),left(nullptr),right(nullptr){}
 };

 void helper(BSTreeNode* root,BSTreeNode*& head, BSTreeNode*& tail){
     if(!root) return;
     helper(root->left,head,tail);
     root->left = tail;
     if(tail){
         tail->right = root;
     }else{
         head = root;
     }
     tail = root;
     helper(root->right,head,tail);
 }

 BSTreeNode* ConvertBSTree2List(BSTreeNode* root){
     BSTreeNode* head = nullptr;//list head
     BSTreeNode* tail = nullptr;//list tail
     helper(root,head,tail);
     return head;
 }

 int main(){
     BSTreeNode* root = new BSTreeNode();
     root->left = new BSTreeNode();
     root->right = new BSTreeNode();
     root->left->right = new BSTreeNode();
     BSTreeNode* head;
     head = ConvertBSTree2List(root);
     return ;
 }