Linked List - Get middle node

"""
https://leetcode.com/problems/middle-of-the-linked-list/description/

Given the head of a singly linked list, return the middle node of the linked list.

If there are two middle nodes, return the second middle node.

2024.08.21 - I think what trips me up about the prompt is the requirement to return the second middle node. It sounds like it'll require some extra logic to handle, when in fact it's just the way things work by default if you use the fast-and-slow-pointer approach.
"""

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        slow = fast = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        return slow