题解 | #求二叉树的层序遍历#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @return int整型二维数组
#
class Solution:
    def levelOrder(self , root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        lay = [[root]]   # 按层存入
        res = []         # 存放结果
        while lay:
            now_lay = lay.pop(0)   # 当前遍历的层
            now_lay_res = []
            next_lay = []          # 下一层的节点
            for tmp in now_lay:
                if tmp:
                    now_lay_res.append(tmp.val)
                    if tmp.left:
                        next_lay.append(tmp.left)
                    if tmp.right:
                        next_lay.append(tmp.right)
            if next_lay: # 如果有元素，存入lay
                lay.append(next_lay)
            res.append(now_lay_res)
        return res