题解 | #求二叉树的层序遍历#

#bfs

# 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]]:
        # write code here
        if root is None: # 判断根节点是否为空
            return
        res = []
        q = [root]
        while q:
            tep = []
            for _ in range(len(q)): # 对于每一层进行遍历
                node = q.pop(0)
                tep.append(node.val) 
                # 判断左右节点是否存在
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            res.append(tep)
        return res