★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Given the root of a binary tree, find the maximum average value of any subtree of that tree.

(A subtree of a tree is any node of that tree plus all its descendants. The average value of a tree is the sum of its values, divided by the number of nodes.) 

Example 1:

Input: [5,6,1]
Output: 6.00000
Explanation: 
For the node with value = 5 we have and average of (5 + 6 + 1) / 3 = 4.
For the node with value = 6 we have and average of 6 / 1 = 6.
For the node with value = 1 we have and average of 1 / 1 = 1.
So the answer is 6 which is the maximum.

Note:

1. The number of nodes in the tree is between 1 and 5000.
2. Each node will have a value between 0 and 100000.
3. Answers will be accepted as correct if they are within 10^-5 of the correct answer.

给你一棵二叉树的根节点 root，找出这棵树的 每一棵 子树的 平均值 中的 最大 值。

子树是树中的任意节点和它的所有后代构成的集合。

树的平均值是树中节点值的总和除以节点数。 

示例：

输入：[5,6,1]
输出：6.00000
解释： 
以 value = 5 的节点作为子树的根节点，得到的平均值为 (5 + 6 + 1) / 3 = 4。
以 value = 6 的节点作为子树的根节点，得到的平均值为 6 / 1 = 6。
以 value = 1 的节点作为子树的根节点，得到的平均值为 1 / 1 = 1。
所以答案取最大值 6。 

提示：

1. 树中的节点数介于 1 到 5000之间。
2. 每个节点的值介于 0 到 100000 之间。
3. 如果结果与标准答案的误差不超过 10^-5，那么该结果将被视为正确答案。

88ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     var ret:Double = 0
16     func maximumAverageSubtree(_ root: TreeNode?) -> Double {
17         if root == nil {return ret}
18         dfs(root)
19         return ret        
20     }
21     
22     func dfs(_ root: TreeNode?) -> (Int,Int)
23     {
24         var cal:Int = 1
25         var sum:Int = root!.val
26         if root!.left != nil
27         {
28             var test:(Int,Int) = dfs(root!.left)
29             cal += test.0
30             sum += test.1
31         }
32         if root!.right != nil
33         {
34             var test:(Int,Int) = dfs(root!.right)
35             cal += test.0
36             sum += test.1
37         }
38         let temp:Double = Double(sum) / Double(cal)
39         if ret < temp
40         {
41             ret = temp
42         }
43         return (cal, sum)
44     }
45 }