递归打印单个链表作业

Question

I have an assignment in my java class that i need to recursively print out a linked list in reverse. I have looked online and found numerous examples of recursive methods that do this but take a node in as a parameter, and to my understanding i need to take in a linked list because i need to print the entire list out. Below is what code i have written, and it works in the sense that it prints out the list, recursively, but it is still in the same order that i created the list. After it prints out the list it throws a no such element exception as well. my main problem/question is wrapping my head around how best to print this recursively.

public void printRecurse2(LinkedList<String> list2)
{
    if(list2 == null)
        return;
    System.out.println(list2.pop());
    printRecurse2(list2);

}

Answer 1

You need to be checking if the node has another next node. This way you defer printing each element until you get to the end of the list:

void printReverse(Node node) {
    if(node.next != null) { // recurse until the last node is found
        printReverse(node.next);  // print the next node first
    }
    System.out.println(node.data); // print out the node(this is only reached after the last node is found
}

This will print the entire list by passing in the first node of your linked list. Then you dont need to pass the entire list to each call. You can also use this to print just part of a list by changing which node you pass to the first call.

Answer 2

Your list will never be null , this is why the recursion doesn't stop in time. You need to check whether the list is not empty, instead of for null, eg

if(list2.peek() == 0)
    return;
System.out.println(list2.pop());
printRecurse2(list2);

This avoids the exception. To print in reverse, you need to start from the other end of the list:

if(list2.peekLast() == null)
    return;
System.out.println(list2.pollLast());
printRecurse2(list2);

Answer 3

As this is homework, I'm only going to guide you, not give you the answer.

In any recursive function, you need at least one base case to end the recursion. You've chosen if (list2 == null) as your base case. At the end of each function call, you're calling printRecurse2(list2) where list2 has had one element popped from it. No matter what, list2 will exist, and not be equal to null . It might , however, be empty . Consider changing your base case.

In regards to the order of items removed, consider the Java API for pop() . Pop removes the first item from the list. So, the first time you call printRecurse2() , the first item will be popped and printed. The next time, the item that was second will now be popped and printed. You may want to look into another method, like removeLast() .

Answer 4

If you need to get the elements in reverse try using the removeLast() method from the LinkedList API, it will return then remove the last element in the LinkedList :

public void printRecurse2(LinkedList<String> list)
{
    if(list.isEmpty()) //base
        return;
    System.out.println(list.removeLast()); //print the last element
    printRecurse2(list); //recurse
}

Additionally, since this method consumes the LinkedList you may want to pass a copy of the LinkedList to this method.