剑指 Offer 32 - I. 从上到下打印二叉树

层次遍历，用到队列（先进先出）.
queue.Queue()
collections.deque()

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from queue import Queue # queue 不是大写
class Solution:
    def levelOrder(self, root: TreeNode) -> List[int]:
        node = [] 
        if root is None: return node # 当输入为[]
        q = Queue()
        q.put(root)
        while not q.empty():
            out = q.get()
            node.append(out.val)
            if out.left is not None: # is not 不是 not
                q.put(out.left)
            if out.right is not None:
                q.put(out.right)
        return node

双端队列

class Solution:
    def levelOrder(self, root: TreeNode) -> List[int]:
        if not root: return []
        res, queue = [], collections.deque()
        queue.append(root)
        while queue:
            node = queue.popleft()
            res.append(node.val)
            if node.left: queue.append(node.left)
            if node.right: queue.append(node.right)
        return res

剑指 Offer 32 - II. 从上到下打印二叉树 II

层次遍历，但要找到每一层最后一个节点
deque操作

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from queue import Queue
from collections import deque
class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root: return []
        node = []
        sub = []
        last_node = root
        q = deque()
        q.append(root)
        while q: # 判断不为空
            out = q.popleft() # 左侧出来
            sub.append(out.val)
            if out.left is not None: # is not 不是 not
                q.append(out.left) 
            if out.right is not None:
                q.append(out.right)
            if out==last_node:
                node.append(sub)
                sub = []
                if q: # 如果没有下一层了
                    last_node = q.pop() # 右侧出来
                    q.append(last_node) # 右侧进去
        return node

循环一层里面的节点，在循环下一层，因为读取第t层，不会含有t+2层

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

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。

只能在加入到列表的时候操作，因为存储的队列要按照二叉树的结构。右边读取队列节点时，总时读最后一个，就不是二叉树结构了。

最后只需要加一个变量记录是奇数还是偶数，然后在sub列表中，左边加入或者右边加入。

# 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 []
        node = []
        sub = deque() # 列表变为双端队列
        last_node = root
        q = deque()
        q.append(root)
        layer = 1
        while q: # 判断不为空
            out = q.popleft() # 左侧出来
            if layer % 2: 
                sub.append(out.val)
            else:
                sub.appendleft(out.val)
            if out.left is not None: # is not 不是 not
                q.append(out.left) 
            if out.right is not None:
                q.append(out.right)
            if out==last_node:
                layer += 1
                node.append(list(sub)) # 把队列换成list 注意可以转换的
                sub = deque()
                if q: # 如果没有下一层了
                    last_node = q.pop() # 右侧出来
                    q.append(last_node) # 右侧进去
        return node