leetcode | 按层访问一棵二叉树 | Python

题目：

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]]

按层访问二叉树，并返回访问结果。

解题思路：

递归+广度优先（BFS）

每个递归中，先得到当前层的所有结点（now）；通过当前层的结点得到数值，保存到结果中（res）；然后根据当前层得到下一层的结点（next_term）；

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def levelOrder(self, root):  #root是树的根节点        """        :type root: TreeNode        :rtype: List[List[int]]        """        if not root:            return []        res = []        self.excute([root],res)        return res            def excute(self,now,res): #递归体，now是当前层的节点，next_term是下一层节点，res是保存得到的结果。        if not now:            return                 tmp = [] #tmp 保存了当前层的数值        next_term = []        for one in now:            tmp.append(one.val)            if one.left:                next_term.append(one.left)            if one.right:                next_term.append(one.right)        res.append(tmp)        self.excute(next_term,res)