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LeetCode-1343. Maximum Product of Splitted Binary Tree

郝春妮 2022-08-10 阅读 82


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:

LeetCode-1343. Maximum Product of Splitted Binary Tree_java

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:

LeetCode-1343. Maximum Product of Splitted Binary Tree_java_02

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;
}
};

 

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