【Description】

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

【Idea】
這個(gè)題也不難， 因?yàn)镺(n)還蠻能打的， 所以po一下
BFS要更好寫一些。 這里每層遍歷node時(shí)直接用了for，沒有用queue，不然還要另起個(gè)變量標(biāo)記

【Solution】

# 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 []
        
        cur = [root]
        res = []
        while cur:
            cur_vals = []
            nextList = []
            for node in cur:
                cur_vals.append(node.val)
                if node.left:
                    nextList.append(node.left)
                if node.right:
                    nextList.append(node.right)
            cur = nextList
            res.append(cur_vals) 
        return res     

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