--- tags: data_structure_python --- # Delete Node in a Linked List <img src="https://img.shields.io/badge/-easy-brightgreen"> 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: ![](https://assets.leetcode.com/uploads/2018/12/28/237_example.png) **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 ### Solution 1: ```python= # 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. """ while node != None: node.val = node.next.val if node.next.next == None: node.next = None node = node.next ``` ### Solution 2: ```python= 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 ```