Question

Formatted question description: https://leetcode.ca/all/1120.html

 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 and 5000.
     Each node will have a value between 0 and 100000.
     Answers will be accepted as correct if they are within 10^-5 of the correct answer.

Algorithm

DFS Define a pair<int, int> for each node, the first position represents the sum of the subtree values rooted at that node, and the second position represents the number of nodes of the subtree; Accumulate these two values of each node from the top; The average number of subtrees is the sum/node, which is stored using a global variable; Use dictionaries for memoized searches to speed up

Code

Java

public class Maximum_Average_Subtree {

    /**
     * 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 {

        double max = Integer.MIN_VALUE;

        public double maximumAverageSubtree(TreeNode root) {
            dfs(root);
            return max;
        }

        // double[]: sum, count
        private double[] dfs(TreeNode root) {
            if (root == null) {
                return new double[]{0, 0};
            }

            double[] l = dfs(root.left);
            double[] r = dfs(root.right);
            double sum = root.val + l[0] + r[0];
            double count = l[1] + r[1] + 1;

            double avg = sum / count;
            if (avg > max) {
                max = avg;
            }

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

// OJ: https://leetcode.com/problems/maximum-average-subtree/
// Time: O(N)
// Space: O(H)
class Solution {
    double ans = 0;
    pair<int, int> dfs(TreeNode *root) {
        if (!root) return {0,0};
        auto left = dfs(root->left), right = dfs(root->right);
        int sum = left.first + right.first + root->val, cnt = left.second + right.second + 1;
        ans = max(ans, (double)sum / cnt);
        return {sum, cnt};
    }
public:
    double maximumAverageSubtree(TreeNode* root) {
        dfs(root);
        return ans;
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maximumAverageSubtree(self, root: TreeNode) -> float:
        def dfs(root):
            if root is None:
                return 0, 0
            ls, ln = dfs(root.left)
            rs, rn = dfs(root.right)
            s = ls + root.val + rs
            n = ln + 1 + rn
            nonlocal ans
            ans = max(ans, s / n)
            return s, n

        ans = 0
        dfs(root)
        return ans

1120 - Maximum Average Subtree

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