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
#
# 用两个列表分别保存当前行和下一行的节点
# 每一层遍历的时候保存节点值
class Solution:
    def levelOrder(self , root: TreeNode) -> List[List[int]]:
        if not root:
            return None
        
        ans = []
        cur = [root]
        nex = []
        nodes = [root]

        while cur:
            temp = []
            for node in cur:    
                temp.append(node.val)
                if node.left:
                    nex.append(node.left)
                if node.right:
                    nex.append(node.right)
            cur = nex
            nex = []
            ans.append(temp)

        return ans