# Merge two linked lists

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.

Return *the head of the merged linked list*.

```
class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
cur = dummy = ListNode()
while list1 and list2:
if list1.val < list2.val:
cur.next = list1
list1, cur = list1.next, list1
else:
cur.next = list2
list2, cur = list2.next, list2
if list1 or list2:
cur.next = list1 if list1 else list2
return dummy.next
```

## Related Problems

Given the head of a linked list, remove the nth node from the end of the list and return its head.

Given the root of a binary tree, flatten the tree into a "linked list":

The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.

The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Given the heads of two singly linked-lists headA and headB, return *the node at which the two lists intersect*. If the two linked lists have no intersection at all, return null.

Given an integer array nums where the elements are sorted in **ascending order**, convert *it to a ***height-balanced ***binary search tree*.