Python function to reverse a linked list

Could anyone please help me to look at what's wrong with the code here?

In order to achieve this:
Define a function that inputs the head node of a linked list, reverses the list and outputs the head node of the reversed list.

Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

class ListNode(object):
    def __init__(self, val: int):
        self.val: int = val
        self.next: Optional[ListNode] = None

class Solution(object):
    def reverseList(self, head):
        if not head:
            return None
        l = []
        while head:
            l.append(head.val)
            head = head.next
        l = l[::-1]
        hn = ListNode(l[0]) 
        for i in range(1, len(l)):
            hn.next = ListNode(l[i])
            nnext = hn.next
            hn = nnext
        return hn

It only returns a single value, what if I want it to output the whole list?
Thank you for your attention and any help with coding/concept/logic would be appreciated.

Creating a Python list kinda defeats the purpose of working with linked lists. you should try to achieve this only by manipulating links between nodes.

For example:

def reverseList(head):
    tail = None
    while head:
        head.next, tail, head = tail, head, head.next
    return tail

If you're going to use a Python list, first fill it with the nodes. Then link each node to the previous one in the list. Finally, return the last node in the list:

def reverseList(head):
    nodes = [None]
    while head:
        nodes.append(head)
        head = head.next
    for i,previous in enumerate(nodes[:-1],1):
        nodes[i].next = previous
    return nodes[-1]

What is wrong with your code is that in creating a new list you start with:

hn = ListNode(l[0])

but inside the loop you lose track of that first node by overwriting hn:

hn = nnext

Finally when you return hn you are referencing the final ListNode created.