LeetCode 剑指 Offer 06. 从尾到头打印链表

题目

输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。

例：
输入：head = [1,3,2]
输出：[2,3,1]

方法一：暴力

遍历链表，将元素值依次存入数组 ans，再将数组中元素翻转

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

class Solution(object):
    def reversePrint(self, head):
        ans = []
        while head != None:
            ans.append(head.val)
            head = head.next
        left, right = 0, len(ans)-1
        while left < right:
            ans[left], ans[right] = ans[right], ans[left]
            left += 1
            right -= 1
        return ans

方法二：递归

先走至链表末端，回溯时依次将节点值加入列表 ，这样就可以实现链表值的倒序输出

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

class Solution(object):
    def reversePrint(self, head):
        return self.reversePrint(head.next) + [head.val] if head else []

参考

代码相关：https://leetcode.cn/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/solution/mian-shi-ti-06-cong-wei-dao-tou-da-yin-lian-biao-d/