LeetCode 2265. Count Nodes Equal to Average of Subtree

原题链接在这里：https://leetcode.com/problems/count-nodes-equal-to-average-of-subtree/description/

题目：

Given the root of a binary tree, return the number of nodes where the value of the node is equal to the average of the values in its subtree.

Note:

• The average of n elements is the sum of the n elements divided by n and rounded down to the nearest integer.
• A subtree of root is a tree consisting of root and all of its descendants. 

Example 1:

Input: root = [4,8,5,0,1,null,6]
Output: 5
Explanation: 
For the node with value 4: The average of its subtree is (4 + 8 + 5 + 0 + 1 + 6) / 6 = 24 / 6 = 4.
For the node with value 5: The average of its subtree is (5 + 6) / 2 = 11 / 2 = 5.
For the node with value 0: The average of its subtree is 0 / 1 = 0.
For the node with value 1: The average of its subtree is 1 / 1 = 1.
For the node with value 6: The average of its subtree is 6 / 1 = 6.

Example 2:

Input: root = [1]
Output: 1
Explanation: For the node with value 1: The average of its subtree is 1 / 1 = 1.

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• 0 <= Node.val <= 1000

题解：

DFS return sum of values in substree and count of nodes in subtree.

when sum/count == root.val, incremental the result by 1.

Time Complexity: O(n). n is the number of nodes in the tree.

Space: O(h). h is the height of the tree.

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int res = 0;
    public int averageOfSubtree(TreeNode root) {
        dfs(root);
        return res;
    }

    private int[] dfs(TreeNode root){
        if(root == null){
            return new int[]{0, 0};
        }

        int[] l = dfs(root.left);
        int[] r = dfs(root.right);
        int sum = l[0] + r[0] + root.val;
        int count = l[1] + r[1] + 1;
        if(sum / count == root.val){
            res++;
        }

        return new int[]{sum, count};
    }
}

类似Maximum Average Subtree.