Force buckle: merging two ordered linked lists

The title is merge 2 An ordered list . The main implementation ideas are as follows ：

1. The linked list is ordered . When comparing, just list1 The value is greater than list2 Of has been inserting data . vice versa .
2. You must use a linked list to get through , You can't use arrays
3. Pay attention to the empty linked list and the termination conditions

# Definition for singly-linked list.
class ListNode:
	def __init__(self, val=0, next=None):
	   	self.val = val
        self.next = next
class Solution:
    def appendList(self,list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        while list2:
            list1.next = ListNode(list2.val)
            list1 = list1.next
            list2 = list2.next

    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        if not list2:
            return list1
        if not list1:
            return list2
        if list1.val > list2.val:
            r = m = ListNode(list2.val)
            list2 = list2.next
            if not list2:
                self.appendList(m,list1)
                return r
        else:
            r = m = ListNode(list1.val)
            list1 = list1.next
        while list1:
            while list1.val > list2.val:
                m.next = ListNode(list2.val)
                m = m.next
                list2 = list2.next
                if not list2:
                    self.appendList(m,list1)
                    return r
            m.next = ListNode(list1.val)
            m = m.next
            list1 = list1.next
        self.appendList(m,list2)
        return r