Maximum Average Subtree

Problem Statement

In the given task, we are provided with a binary tree originating from a single root node. Our objective is to determine the maximum average value from any of the subtrees in this binary tree. For the purpose of this problem, a subtree is defined as any node in the tree, including all of its descendants. The average of a subtree is calculated by summing up all the node values in the subtree and then dividing by the total number of nodes in that subtree. The acceptable error margin for the answer is within (10^{-5}) of the true value.

Examples

Example 1

Input:

root = [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.

Example 2

Input:

root = [0,null,1]

Output:

1.00000

Constraints

• The number of nodes in the tree is in the range [1, 104].
• 0 <= Node.val <= 105

Approach and Intuition

Figuring out the maximum average value of a subtree involves examining averages for different possible subtrees:

1. Definition of a Subtree:

   • Any node of the tree can be the root of a subtree, encompassing itself and its descendants.

2. Calculation of Average:

   • For each subtree rooted at a node, calculate the sum of its values and the count of its nodes to obtain the average for that particular subtree.

3. Recursive Strategy:

   • Utilize a recursive function to navigate down the tree. At each node, consider it as a potential root of a subtree and calculate the average for it. Recurse for its left and right children. This function should return both the sum and count of nodes for any node’s subtree which can be used by its parent node calculations.

4. Storing Maximum:

   • During the recursive traversal, continuously compare and update a global variable that tracks the maximum average found among all checked subtrees.

5. Complexity Consideration:

   • Given that recursion will reach out to every node and potentially calculate sums and counts, the time complexity will primarily be (O(n)), where (n) is the number of nodes in the tree. The use of recursion will call for additional space in the form of recursion stack, which in the worst case (unbalanced tree) could also be (O(n)).

By following these systematic steps, using efficient computation ways and by maintaining current subtree sums and counts, the solution ensures that it scales according to constraints and processes possibly large trees effectively. The checks against edge conditions (like trees with a single node or with large values) are implicit in the straightforward but thorough recursive computation and comparison.

Solutions

class Solution {
public:
    double findMaxAverageSubtree(TreeNode* root) {
        return computeStats(root).maxAverage;
    }
    
private:
    struct Stats {
        int totalNodes;
        int totalSum;
        double maxAverage;
    };
    
    Stats computeStats(TreeNode* root) {
        if (!root) return {0, 0, 0};
    
        Stats leftStat = computeStats(root->left);
        Stats rightStat = computeStats(root->right);
    
        int totalNodes = leftStat.totalNodes + rightStat.totalNodes + 1;
        int totalSum = leftStat.totalSum + rightStat.totalSum + root->val;
        double maxAve = max(
                (double(totalSum) / totalNodes),
                max(leftStat.maxAverage, rightStat.maxAverage)
        );
    
        return {totalNodes, totalSum, maxAve};
    }
};