Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.
Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]
递归法
通过递归的方法去删除节点
递归程序会先一路遍历来到节点尾部
从后往前把val符合的节点进行删除, 并重新把链表连接起来
class Solution:def removeElements(self, head: ListNode, val: int) -> ListNode:if head is None:return head# removeElement方法会返回下一个Node节点head.next = self.removeElements(head.next, val)if head.val == val:next_node = head.next else:next_node = headreturn next_node