题解 | #删除链表的节点#

构建虚拟头节点dummy，遍历链表即可

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param val int整型 
# @return ListNode类
class Solution:
    def deleteNode(self , head: ListNode, val: int) -> ListNode:
        # write code here
        if not head:
            return None
        dummy = ListNode(-1)
        dummy.next = head
        tmp = dummy
        while head:
            if head.val == val:
                tmp.next = head.next
                tmp = head.next
            tmp = head
            head = head.next
        return dummy.next