LeetCode-1343. Maximum Product of Splitted Binary Tree

LeetCode-1343. 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: 110Explanation: 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: 90Explanation: 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; }};

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