1339. Maximum Product of Splitted Binary Tree

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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 10^9 + 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)

Example 3:

Input: root = [2,3,9,10,7,8,6,5,4,11,1]

Output: 1025

Example 4:

Input: root = [1,1]

Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 5 * 10^4].
• 1 <= Node.val <= 10^4

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解法1

/**
 * 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 {
    
    long res = 0, total = 0, sub = 0;
    
    public int maxProduct(TreeNode root) {
        // 得到所有节点的值的总和
        total = helper(root); 
        
        // 计算最大乘积
        helper(root);
        
        return (int)(res % (int)(1e9 + 7));
    }

    private long helper(TreeNode root) {
        if (root == null) return 0;
        // 当前node以及子节点的和
        sub = root.val + helper(root.left) + helper(root.right);
        
        // 把当前点和父节点断开, 计算乘积
        res = Math.max(res, sub * (total - sub));
        
        return sub;
    }
}