Maximum Product of Splitted Binary Tree

Problem Statement

In this problem, you are given a binary tree with the root as its primary node. Your task is to determine how to split this tree into two distinct subtrees by severing one edge, and then calculate the product of the sum of the node values in each subtree. Your goal is to maximize this product. Given the nature of large numbers, you must return the result as modulo (10^9 + 7). It's essential to maximize the product before taking the modulo operation. This ensures that the computation deals with large values effectively while abiding by the constraints imposed by large-scale calculations.

Examples

Example 1

Input:

root = [1,2,3,4,5,6]

Output:

110

Explanation:

Remove the red edge and get 2 binary trees with sum 11 and 10. Their product is 110 (11*10)

Example 2

Input:

root = [1,null,2,3,4,null,null,5,6]

Output:

90

Explanation:

Remove the red edge and get 2 binary trees with sum 15 and 6.Their product is 90 (15*6)

Constraints

• The number of nodes in the tree is in the range [2, 5 * 104].
• 1 <= Node.val <= 104

Approach and Intuition

1. Tree Traversal for Sum Calculation:

   • To determine the best way to split the tree, you first need to understand the total sum of all nodes in the tree. A recursive approach (such as a post-order traversal) can efficiently calculate this. This traversal helps gather information about each subtree rooted at every node.

2. Splitting the Tree:

   • After determining the total sum of the tree, the next step is to explore potential split points. As you traverse the tree (again, preferably in a post-order fashion), compute the sum of each subtree.
   • Every node visited represents a potential split point where its subtree sum and the sum of the rest of the tree can form the two segments needed.

3. Maximizing the Product:

   • For each split, calculate the product of the sum of the current subtree and the sum of the rest of the tree (total sum minus current subtree sum).
   • Keep track of the maximum product found during these calculations.

4. Return Statement with Modulo Operation:

   • After traversing all potential splits and computing the respective products, the final step is to return the highest product modulo (10^9 + 7). This ensures that even with a maximum product computation, the output remains manageable and within given constraints.

Intuitive Insights:

• Larger Splits: Often, balanced splits or near-balanced splits (where the tree is divided into two nearly equal sum subtrees) might offer higher products due to the multiplicative property. An entirely skewed split (one very small subtree and one large subtree) usually yields a smaller product.
• Tree Configuration: The structure of the tree can significantly affect which splits offer the maximum product—balanced trees contrast with highly skewed trees (like linked lists) in their splitting potential.

By using an efficient tree traversal method and maintaining a keen eye on subtrees' sum during the traversal, one can effectively solve this problem by evaluating the strategic points of tree division.

Solutions

class BinaryTreeMaxProduct {

    private static final int MODULO = 1000000007;

    private List<Integer> subtreeSums = new ArrayList<>();

    public int maximalProduct(TreeNode root) {
        int completeSum = calculateTreeSum(root);
        int closestSum = 0;
        int minimalDistance = Integer.MAX_VALUE;
        for (int partSum : subtreeSums) {
            int distance = Math.abs(completeSum - 2 * partSum);
            if (distance < minimalDistance) {
                minimalDistance = distance;
                closestSum = partSum;
            }
        }
        return moduloProduct(closestSum, completeSum - closestSum, MODULO);
    }

    private int moduloProduct(int x, int y, int mod) {
        int res = 0;
        int term = x;
        while (y > 0) {
            if ((y & 1) == 1) {
                res = (res + term) % mod;
            }
            term = (term * 2) % mod;
            y >>= 1;
        }
        return res;
    }

    private int calculateTreeSum(TreeNode node) {
        if (node == null) return 0;
        int sumLeft = calculateTreeSum(node.left);
        int sumRight = calculateTreeSum(node.right);
        int total = sumLeft + sumRight + node.val;
        subtreeSums.add(total);
        return total;
    }
}