Final answer:
The question asks for a recursive C++ function to display linked list nodes except for the first node. The provided example defines a base case to handle the end of the list or first node, and a recursive case to print and proceed through the rest of the list.
Step-by-step explanation:
The question pertains to writing a recursive function in C++ that displays all nodes of a linked list except for the first node. When using recursion, the base case for our function will be to check if the pointer to the linked list is null or if it's the first node, and the recursive case will involve moving to the next node and displaying its value.
Example Recursive Function
To accomplish this, consider the following code example:
void displayExceptFirst(ListNode* head) {
if (head == nullptr || head->next == nullptr) // Base case: end of list or first node
return;
else {
cout << head->next->data << " "; // Display data of current node
displayExceptFirst(head->next); // Recursive call
}
}
To use this function, simply call displayExceptFirst with the head of your linked list as an argument. The function will skip the first node, then print and recursively continue through the remaining nodes in the list.