May 9, 2022
diff --git a/python3/21.py b/python3/21.py

 # https://leetcode.com/problems/merge-two-sorted-lists/
 # Iterative solution
 class Solution:
    def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        tempHead = ListNode() # Empty node to be head of merged list
        temp = tempHead # End node of merged list, append new nodes to end of temp

        # While BOTH l1 and l2 still have nodes
        while (l1 is not None and l2 is not None):
            # If current node of l1 has a smaller value than current node of l2
            if (l1.val < l2.val):
                # Append l1 node to merged list
                temp.next = l1
                # Go to next l1 node
                l1 = l1.next
            else:
                temp.next = l2
                l2 = l2.next

            # Go to new end of merged list
            temp = temp.next

        # Either l1, l2, or neither have nodes left
        # Can directly append all of the remaining nodes to merged list
        temp.next = l1 if l1 is not None else l2

        # tempHead is empty node, return the actual head node of merged list
        return tempHead.next
Original file line number	Diff line number	Diff line change
		@@ -0,0 +1,30 @@

		# https://leetcode.com/problems/merge-two-sorted-lists/
		# Iterative solution
		class Solution:
		def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
		tempHead = ListNode() # Empty node to be head of merged list
		temp = tempHead # End node of merged list, append new nodes to end of temp

		# While BOTH l1 and l2 still have nodes
		while (l1 is not None and l2 is not None):
		# If current node of l1 has a smaller value than current node of l2
		if (l1.val < l2.val):
		# Append l1 node to merged list
		temp.next = l1
		# Go to next l1 node
		l1 = l1.next
		else:
		temp.next = l2
		l2 = l2.next

		# Go to new end of merged list
		temp = temp.next

		# Either l1, l2, or neither have nodes left
		# Can directly append all of the remaining nodes to merged list
		temp.next = l1 if l1 is not None else l2

		# tempHead is empty node, return the actual head node of merged list
		return tempHead.next