面试题06. 从尾到头打印链表

https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/

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）。

示例 1：

输入：head = [1,3,2]

输出：[2,3,1]

限制：

0 <= 链表长度 <= 10000

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小技巧

从尾到头打印节点，符合 后进先出 的特点:

• 借助 栈 结构来实现

• 通过天然的栈结构 递归 来实现，链表过长时，可能会导致栈溢出

备注

• 时间复杂度：O(n)

• 空间复杂度：O(n)，栈或者递归所需

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方法一：递归

# 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):
        """
        :type head: ListNode
        :rtype: List[int]
        """
        def recur(node):
            # terminator
            if not node:
                return
            # drill down
            recur(node.next)

            # process current level
            self.rs.append(node.val)
        self.rs = []
        recur(head)
        return self.rs

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方法二：非递归(栈)

# 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):
        """
        :type head: ListNode
        :rtype: List[int]
        """
        stack = []
        while head:
            stack.append(head.val)
            head = head.next
        return stack[::-1]