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：

 1 # Definition for a binary tree node.
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 
 8 class Solution:
 9     def averageOfLevels(self, root: TreeNode) -> List[float]:
10         result = []
11         self.BFS(root,result)
12         return result
13         
14         
15     def BFS(self,root,result):
16         if root == None: return
17         queue = [root]
18         while queue:#如果queue不为空
19             val = [i.val for i in queue if i]
20             if len(val):result.append(sum(val)/len(val))
21             queue = [child for node in queue if node for child in (node.left,node.right)]
22

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

posted on 2019-09-18 10:38 狂奔的蜗牛163 阅读(237) 评论(0) 编辑收藏举报