[剑指Offer-06][简单] 从尾到头打印链表
题目描述
输入:head = [1,3,2]
输出:[2,3,1]解题思路
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
return self.reversePrint(head.next) + [head.val] if head is not None else []最后更新于