# Add two numbers - linked list representation of numbers

You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse order**, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

```
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
dummyHead = ListNode(0)
tail = dummyHead
carry = 0
while l1 is not None or l2 is not None or carry != 0:
digit1 = l1.val if l1 is not None else 0
digit2 = l2.val if l2 is not None else 0
sum = digit1 + digit2 + carry
digit = sum % 10
carry = sum // 10
newNode = ListNode(digit)
tail.next = newNode
tail = tail.next
l1 = l1.next if l1 is not None else None
l2 = l2.next if l2 is not None else None
result = dummyHead.next
dummyHead.next = None
return result
```

## Related Problems

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*.

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 a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

Given the roots of two binary trees p and q, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.