1343. Maximum Product of Splitted Binary Tree
https://leetcode.com/problems/maximum-product-of-splitted-binary-tree/
Given a binary tree root
. Split the binary tree into two subtrees by removing 1 edge such that the product of the sums of the subtrees are maximized.
Since the answer may be too large, return it modulo 10^9 + 7.
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
二叉树一条边能把树分成两棵,问能任意去掉一条边,两棵树各自的和相乘最大会是多少。遍历统计出整棵树的和,再遍历每个结点统计出该结点的子树和是多少,再看当前子树的和 * (sum - 当前子树和)是否是最大解。也可以用集合统计出所有可能的和,再分别看它们所形成的乘积最大是多少。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
const unsigned int M = 1000000007;
long res = 0;
int sum(TreeNode *node, long total = 0) {
if (node == nullptr) return 0;
int l = sum(node->left, total);
int r = sum(node->right, total);
const int s = l + r + node->val;
res = max(res, s * (total - s));
return s;
}
int maxProduct(TreeNode* root) {
const int total = sum(root);
sum(root, total);
return res % M;
}
};
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