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配合

# Definition for a binary tree node.# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    def levelOrder(self, root: TreeNode) -> List[List[int]]:        if not root:            return []        queue = [root]        res = []        while queue:            tmp = []    # 存储单层的结果            for _ in range(len(queue)):                node = queue.pop(0)	# 队首元素                tmp.append(node.val)                if node.left:                    queue.append(node.left)                if node.right:                    queue.append(node.right)                                res.append(tmp)        return res

链接：https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof/solution/mian-shi-ti-32-ii-cong-shang-dao-xia-da-yin-er-c-5/

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