Swapping adjacent elements of linked list

Question

The below is my code to recursive swap the adjacent elements of a linked list. I am losing the pointer to every second element after the swap. The input is 1->2->3->4->5->6-

Answer 1

In fact it is enough to swap only the data members of nodes. There is no need to swap the pointers themselves.

Nevertheless if to use your approach then the function can look like

void SwapList( node *head )
{
    if ( head != nullptr && head->next != nullptr )
    {
        node *next = head->next;
        std::swap( *head, *next );
        std::swap( head->next, next->next );

        SwapList( head->next->next );
    }
}

Here is a demonstrative program

#include 
#include 

struct node
{
    int value;
    node *next;
};

node * AddNode( node *head, int value )
{
    head = new node { value, head };

    return head;
}

void PrintList( node *head )
{
    for ( ; head != nullptr; head = head->next )
    {
        std::cout << head->value << ' ';
    }
}

void SwapList( node *head )
{
    if ( head != nullptr && head->next != nullptr )
    {
        node *next = head->next;
        std::swap( *head, *next );
        std::swap( head->next, next->next );

        SwapList( head->next->next );
    }
}

int main() 
{
    node *head = nullptr;

    for ( int i = 10; i != 0; )
    {
        head = AddNode( head, --i );
    }

    PrintList( head );
    std::cout << std::endl;

    SwapList( head );

    PrintList( head );
    std::cout << std::endl;

    return 0;
}

The output is

0 1 2 3 4 5 6 7 8 9 
1 0 3 2 5 4 7 6 9 8 

You can use the shown function as a template (or base) for your function.