使用 C++ 递归打印 LinkedList

问题描述

我正在尝试创建一个函数来递归地打印出我的链接列表，但我在这样做时遇到了麻烦，因为递归太难了.

I'm trying to create a function that would print out my link list recursively, but I'm having trouble doing that, because recursion is just hard.

这是我写的函数，明明是带参数的，但是不知道怎么传.并且可能输出是错误的.

This is the function I wrote, obviously takes a parameter, but I don't know how to pass it. And probably the output is wrong.

我使用了 typedef:

I used typedef:

 typedef struct node* nodePtr;

多亏了其中一个人的输入，我更新了我的函数，看起来像这样，但现在 Visual Studio 给出了一个错误:

and thanks to the input from one of the guys, I updated my function to look like this, but now visual studio is giving an error that says:

声明与 void List::PrintListRecursively 不兼容"，所以我想知道我传递参数的方式只是略有不同.

"Declaration is incompatible with void List::PrintListRecursively", so I wonder that the way I pass the parameter is just a slight different.

先谢谢你

void List::PrintListRecursively(nodePtr curr ){

    if (curr==NULL)
    {
        cout << "
";
        return;
    }
    cout << curr->data <<endl;
    PrintListRecursively(curr->next);


}

我没有递归地编写了相同的函数:

I wrote the same function not recursively:

void List::PrintList(){
    curr = head;
    while(curr != NULL)
    {
        cout << curr->data <<endl;
        curr = curr->next;
    }
}

这个效果很好.有人可以帮忙解决递归部分并帮助我找出问题所在.不要太刻薄.

and this one works great. Could somebody help out with the recursion part and help me find out whats wrong. Don't be too mean.

推荐答案

您的递归版本需要输入:

Your recursive version needs an input:

void List::PrintListRecursively(Node* curr)
{
    if (curr==NULL)
    {
        cout << "
";
        return;
    }
    cout << curr->data <<endl;
    PrintListRecursively(curr->next);
}

然后你会使用头指针调用它:

Which you would then call using the head pointer:

list.PrintListRecursively(list.GetHead());

或者你可以创建一个不带参数的版本:

Or you could create a version that takes no parameters:

void List::PrintListRecursively()
{
    PrintListRecursively(GetHead());
}

调用带指针参数的版本.

Which calls the version that takes the pointer parameter.

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