void del(node* &head, int val)
{
    if (head == NULL) {
        cout << "Element not present in the list\n";
        return;
    }
    if (head->info == val) {
        node* t = head;
        head = head->link;
        delete (t);
        return;
    }
    del(head->link, val);
}

Intuition:

• We are passing node* (node pointer) as a reference to the 'del' function (as in node* &head).

• Suppose node containing 'val' is the first node, head = head->next changes the actual head in the main function which now points to the current beginning of the list (which was second node before deletion).

• Deleting any other node, now since current node* is derived from previous node's next and previous node's next is passed by reference , thus we can say that the pointer being on the current node can alter previous node's next.

• Thus when we do head = head->next means previous node's next gets changed to head->next which is the required operation while deletion of a node in the linked list.

• Thus there is no need to have a separate case for deletion of first node or last node.