如何返回链表的中间节点

Question

I'm confused about how the while loop condition works in a linked list when checking for the middle node of a linked list

This is the correct code that I have for finding the middle node of a linked list

class Node(object):
    def __init__(self, data):
        self.data = data
        self.next = None


class linkedList(object):
    def __init__(self):
        self.head = None

    def append(self, data):
        node = Node(data)
        if self.head == None:
            self.head = node
        temp = self.head
        while temp.next:
            temp = temp.next
        temp.next = node

    def middle(self):
        first = self.head
        second = self.head
        while second and second.next:
            second = second.next.next
            first = first.next
        print(first.data)

If I change the while loop to

while second:

or

while second.next:

I receive an error that says

AttributeError: 'NoneType' object has no attribute 'next' on line 24

I'm just wondering why it's important to have both second and second.next

Answer 1

The solution works by using two pointers. First one takes 1 step at a time and second 2 steps at a time. However, in taking 2 steps there are two things to verify:

1. There is a next step that is valid
2. There is a step after the next step mentioned above

If you skip one check it will enter the loop but at the boundary condition it not find the next. To illustrate: If there are 4 nodes and you only check for second.next at the 3nd node you'll have second.next as valid and you'll enter the while loop but inside that you are directly accessing second.next.next

F,S | 1 --> 2 --> 3 --> 4 --> None

Answer 2

For starters, your append method doesn't work, and gets stuck in an infinite while loop, since you don't exit out of append on adding the first element. The correct version is

def append(self, data):
    node = Node(data)
    if self.head == None:
        self.head = node
        return
    else:
        temp = self.head
        while temp.next:
            temp = temp.next
        temp.next = node

As to your other question, we want to find the middle of the loop for both even and odd lists, the second.next covers the odd list case, and the second covers the even list case, since either the 2nd pointer will point to null, or it will be null itself, and if you use only one of them, you will get the error you described it, hence you need to have both conditions in the while loop