Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list – head = [4,5,1,9], which looks like following:

**Example 1**:

Input: head = [4,5,1,9], node = 5

Output: [4,1,9]

Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

**Example 2**:

Input: head = [4,5,1,9], node = 1

Output: [4,5,9]

Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

**Note**:

- The linked list will have at least two elements.
- All of the nodes’ values will be unique.
- The given node will not be the tail and it will always be a valid node of the linked list.
- Do not return anything from your function.

## Solution

```
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def deleteNode(self, node: ListNode) -> None:
if node.next:
node.val = node.next.val
node.next = node.next.next
```

## Test Cases

```
test = Solution()
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
test.deleteNode(head)
assert head.val == 2
assert head.next.val == 3
assert head.next.next.val == 4
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
test.deleteNode(head.next)
assert head.val == 1
assert head.next.val == 3
assert head.next.next.val == 4
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
test.deleteNode(head.next.next)
assert head.val == 1
assert head.next.val == 2
assert head.next.next.val == 4
head = ListNode(1)
head.next = ListNode(2)
test.deleteNode(head)
assert head.val == 2
print('All Passed!')
```

## Big O Analysis

**Space Complexity**: O(1)

**Time Complexity**: O(1)