python 从尾到头打印链表

从尾到头打印链表

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]

提供三种题解

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

class Solution:
    """
    解题思路一: 递归解法
    1.f(-r) = f(head.next) + head.val
    2.f(head.next) 为递归函数

    解题思路二: 利用栈的特性，先入后出
    1.遍历链表，不停的入栈
    2.不停的弹出栈，并把值放到数组中

    解题思路三: 
    1.先翻转整个链表
    2.再顺序输出
    """

    # 解题思路一
    def reversePrint(self, head: ListNode) -> List[int]:
        result = list()
        self.reverseTrave(head, result)
        return result

    def reverseTrave(self, head, result):
        if not head:
            return 
        self.reverseTrave(head.next, result)
        result.append(head.val)

    # 解题思路二
    def reversePrint(self, head: ListNode) -> List[int]:
        stack = list()

        result = list()
        # 遍历入栈
        while head:
            stack.append(head.val)
            head = head.next
         
         # 遍历出栈
        while stack:
            result.append(stack.pop())
        
        return result

    # 解题思路三
    def reversePrint(self, head: ListNode) -> List[int]:
        new_head = None

        result = list()

        # 翻转链表
        while head:
            per = head.next
            head.next = new_head
            new_head = head
            head = per

        # 遍历翻转后的链表
        while new_head:
            result.append(new_head.val)
            new_head = new_head.next
        
        return result