[LeetCode] 1120. Maximum Average Subtree
作者:互联网
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 an average of (5 + 6 + 1) / 3 = 4. For the node with value = 6 we have an average of 6 / 1 = 6. For the node with value = 1 we have an average of 1 / 1 = 1. So the answer is 6 which is the maximum.
Note:
- The number of nodes in the tree is between
1
and5000
. - Each node will have a value between
0
and100000
. - Answers will be accepted as correct if they are within
10^-5
of the correct answer.
子树的最大平均值。
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。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/maximum-average-subtree
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既然是计算子树的信息,那么思路应该是后序遍历。这里的helper函数有一些特别,需要记录两个信息,一个是所有子树值的和,一个是子树节点的个数。这里我用一个长度为2的数组记录这两个信息。
时间O(n)
空间O(n)
Java实现
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 double res = 0; 18 19 public double maximumAverageSubtree(TreeNode root) { 20 if (root == null) { 21 return res; 22 } 23 helper(root); 24 return res; 25 } 26 27 private int[] helper(TreeNode root) { 28 int[] arr = new int[2]; 29 if (root == null) { 30 return arr; 31 } 32 int[] left = helper(root.left); 33 int[] right = helper(root.right); 34 //设置当前树的元素和 35 arr[0] = left[0] + right[0] + root.val; 36 //设置节点个数 37 arr[1] = left[1] + right[1] + 1; 38 //更新平均值 39 res = Math.max(res, (double) arr[0] / arr[1]); 40 return arr; 41 } 42 }
标签:right,TreeNode,val,1120,Average,Subtree,int,root,left 来源: https://www.cnblogs.com/cnoodle/p/14414330.html