使用递归打印链表元素

Question

我在Hackerrank上解决反向挑战的指纹

void ReversePrint(node*head)方法接受一个参数-链表的头部。您不应该从stdin/console中读取任何输入。头部可能是空的，所以不应该打印任何东西。按照与stdout/console相反的顺序打印链表的元素（使用printf或cout），每行一个。

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    #include <vector>
    void ReversePrint(Node *head)
{
  // This is a "method-only" submission. 
  // You only need to complete this method. 

    std::vector<int> nodeList;
    if(head != NULL){

        while(head != NULL){
            nodeList.push_back(head->data);
            head = head->next;            
        }

        for (std::vector<int>::iterator it = nodeList.end()-1 ; it != nodeList.begin()-1; --it){
            std::cout << *it <<endl;
       }
    }

}

std::vector<int> nodeList;
void ReversePrint(Node *head){
    if(head != NULL){
        nodeList.push_back(head->data);
        ReversePrint(head->next);
    }
    else{
        for (std::vector<int>::iterator it = nodeList.end()-1 ; it != nodeList.begin()-1; --it){
            std::cout << *it <<endl;
       }

    }

}

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NB：节点的结构为struct Node{int data；struct Node*next；}

Answer 1

为什么这么复杂？

/* Function to reverse print the linked list */
void ReversePrint(Node* head)
{
    // Base case  
    if (head == NULL)
       return;

    // print the list after head node
    ReversePrint(head->next);

    // After everything else is printed, print head
    std::cout << head->data << '\n';
}