LeetCode-剑指 Offer 32 - III. 从上到下打印二叉树 III

请实现一个函数按照之字形顺序打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右到左的顺序打印，第三行再按照从左到右的顺序打印，其他行以此类推。

例如:
给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其层次遍历结果：

[
  [3],
  [20,9],
  [15,7]
]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof
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思路1：和上一道基本一样，就是多设置一个flag变量，循环执行append方法和insert方法，flag通过=not flag来实现交替执行上述两种方法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root: return []
        queue, res, flag  = collections.deque(), [], True
        queue.append(root)
        while queue:
            temp = []
            for _ in range(len(queue)):
                out = queue.popleft()
                if not flag: 
                    temp.insert(0, out.val)
                else:
                    temp.append(out.val)
                if out.left: queue.append(out.left)
                if out.right: queue.append(out.right)
            flag = not flag
            res.append(temp)
        return res

思路2：用了 collections.deque()的appendleft新方法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root: return []
        queue, res = collections.deque([root]), []
        while queue:
            temp = collections.deque()
            for _ in range(len(queue)):
                out = queue.popleft()
                if len(res) % 2: temp.appendleft(out.val)
                else: temp.append(out.val)
                if out.left: queue.append(out.left)
                if out.right: queue.append(out.right)
            res.append(list(temp))
        return res