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(10);
    root->left = new BSTreeNode(6);
    root->right = new BSTreeNode(14);
    root->left->right = new BSTreeNode(8);
    BSTreeNode* head;
    head = ConvertBSTree2List(root);
    return 0;
}