Daily LeetCode - by Vítor Belim

Description

Given the root of a binary tree, split the binary tree into two subtrees by removing one edge such that the product of the sums of the subtrees is maximized.

Return the maximum product of the sums of the two subtrees. Since the answer may be too large, return it modulo 10⁹ + 7.

Note that you need to maximize the answer before taking the mod and not after taking it.

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

Solutions

This is the cleanest variant of the two-pass approach for splitting a binary tree into two pieces with the maximum product of their sums. The first dfs pass turns every node.val into the sum of its entire subtree and saves the grand total. The key simplification is in dfs2: instead of looking down at each node's children, it evaluates the cut at the node itself — detaching the subtree rooted at node leaves two pieces summing to node.val and total - node.val, so a single expression (total - node.val) * node.val covers every possible split as the traversal visits each node. Checking the root this way is harmless, since cutting "above" the root leaves an empty piece and a product of 0. The largest product found is returned modulo 10^9 + 7.

Language: javascript(2026-01-07 10:36)DONE

CPU Performance100.00%

Memory Performance100.00%

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxProduct = function(root) {
    const dfs = (node) => {
        if (!node) return 0;

        node.val += dfs(node.left) + dfs(node.right);

        return node.val;
    };

    const total = dfs(root);
    const mod = 10 ** 9 + 7;
    let max = 0;

    const dfs2 = (node) => {
        if (!node) return;

        max = Math.max(max, (total - node.val) * node.val);

        dfs2(node.left);
        dfs2(node.right);
    };

    dfs2(root, 0);

    return max % mod;
};

Description

Solutions

AI Explanation