Delete Node in a Linked List

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:

    4 -> 5 -> 1 -> 9

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.

My solution:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        node.val = node.next.val
        node.next = node.next.next

This is good description.

I was hope to understand it by walking through it step by step and getting feedback. Can any confirm my description below is correct.

class Solution(object):
    def deleteNode(self, node):
        
        # Summary: To achieve appearance of deleting the node, we actually need to copy the data for the next node 
        # into the node that we were asked to delete then update the node's pointer to the next nodes pointer value
        
        # Input third node with value 3
        #   {value: 3, pointer: fourth node}

        # Debugging 
        node.val = node.next.val
        #   node.val equals 3
        #   node.next.val equals 4
        #   node.val assigned the value of 4
        # Output
        #   1 -> 2 -> 4 -> 4
        
        node.next = node.next.next
        # node.next = third node 
        # node.next.next = NULL/None
        # node.next assigned NULL/None
        # Output
        #   1 -> 2 -> 4