Python3解leetcode Average of Levels in Binary Tree

问题描述：

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

Input:
    3
   / 
  9  20
    /  
   15   7
Output: [3, 14.5, 11]
Explanation:
The average value of nodes on level 0 is 3,  on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].

思路：

考虑BFS算法，因为这是第一次碰到BFS算法，因而将该题记录

代码：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def averageOfLevels(self, root: TreeNode) -> List[float]:
        result = []
        self.BFS(root,result)
        return result
        
        
    def BFS(self,root,result):
        if root == None: return
        queue = [root]
        while queue:#如果queue不为空
            val = [i.val for i in queue if i]
            if len(val):result.append(sum(val)/len(val))
            queue = [child for node in queue if node for child in (node.left,node.right)]
 

BFS的基本思路是：将每一层的结点放置于一个list中，然后遍历List对每个结点进行相应操作，下一步更新该list，将该list放置更下一层的结点。该算法不用递归调用自身，相对而言理解比较容易