reverse link list using recursion

I am trying to reverse a link list using recursion. I have written a function "reverse(node* ptr) for this I am getting output as 40 40 20 while i expect output to be 40 , 20 , 10. below is the code posted.

class list {
    //some code;

    void reverse()
    {
        node* temp = new node;
        temp =first;
        reverse(temp);
     开发者_如何学编程   temp =NULL;
        delete temp;
    }

    void reverse(node* ptr) {


        if(ptr->next != NULL)
        {
           ptr =ptr->next;
           reverse(ptr);
        }
        cout << ptr->data << endl;
    }
    // some code;
};

int main()
{
    list ll;
    ll.insert(18);
    ll.insert(20);
    ll.insert(40);
    ll.display();
    ll.reverse();
    return 0;
}

please suggest what I am doing wrong here.

Thanks

Before even talking about the linked list, there is a major problem with your code:

void reverse()
{
    node* temp = new node;
    temp =first;
    reverse(temp);
    temp =NULL;
    delete temp;
}

You allocate space for a node and then change what it is pointing to, to first. This means the memory you allocated will be leaked. Not only that but you then set it to NULL and then try to free it. You can't free NULL!

I believe you simply meant:

void reverse()
{
    reverse(first);
}

Simple. On to the linked list:

if(ptr->next != NULL)
{
    ptr =ptr->next;
    reverse(ptr);
}

You set ptr to the next element so when reverse() returns it will be one element ahead. I believe you meant:

if(ptr->next != NULL)
{
    reverse(ptr->next);
}

You should get rid of the line ptr =ptr->next;.

The main objective is to print all the nodes that come after the current node before printing value of the current node. So simply a call to reverse(ptr->next) followed by a cout<<ptr->data<<endl should suffice, since the first call takes care of all nodes after ptr. There is no need to advance the pointer as we want to print the current node at the end.