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

Constraints:

• Each tree has at most 50000 nodes and at least 2 nodes.
• Each node's value is between [1, 10000].

题解：

class Solution {
public:
    int mod = 1e9 + 7;
    void calSum(TreeNode* &root) {
        if (root != NULL) {
            calSum(root->left);
            calSum(root->right);
            if (root->left != NULL) {
                root->val += root->left->val;
            }
            if (root->right != NULL) {
                root->val += root->right->val;
            }
        }
    }
    void getMax(TreeNode* &root, int sum, long long &res) {
        if (root != NULL) {
            getMax(root->left, sum, res);
            getMax(root->right, sum, res);
            if (root->left != NULL) {
                long long val = (long long)(sum - root->left->val) * (long long)root->left->val;
                res = max(res, val);
            }
            if (root->right != NULL) {
                long long val = (long long)(sum - root->right->val) * (long long)root->right->val;
                res = max(res, val);
            }
        }
    }
    int maxProduct(TreeNode* root) {
        long long res = 0;
        if (root == NULL) {
            return 0;
        }
        calSum(root);
        getMax(root, root->val, res);
        return res % mod;
    }
};