1339.Maximum Product of Splitted Binary Tree

tags: Medium,Tree,Binary Tree,DFS

1339. Maximum Product of Splitted Binary Tree

題目描述

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 10⁹ + 7.

Note that you need to maximize the answer before taking the mod and not after taking it.

範例

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)

Constraints:

• The number of nodes in the tree is in the range [2, 5 * 10⁴].
• 1 <= Node.val <= 10⁴

解答

C++

class Solution {
public:
    long long result, total;
    long long dfs(TreeNode* root) {
        if (!root) return 0;
        if (root->left == root->right) return root->val;
        long long l_sum = dfs(root->left);
        long long r_sum = dfs(root->right);
        result = max(result, (total - l_sum) * l_sum);
        result = max(result, (total - r_sum) * r_sum);
        return root->val + l_sum + r_sum;
    }
    int maxProduct(TreeNode* root) {
        total = dfs(root);
        dfs(root);
        return result % 1000000007;
    }
};

Yen-Chi ChenSat, Dec 10, 2022

Python

class Solution:
    def maxProduct(self, root: Optional[TreeNode]) -> int:
        all_sums = []
        def get_tree_sum(root):
            if not root: return 0

            left_sum = get_tree_sum(root.left)
            right_sum = get_tree_sum(root.right)
            total_sum = root.val + left_sum + right_sum

            all_sums.append(total_sum)
            return total_sum

        best = 0
        total = get_tree_sum(root)
        for s in all_sums:
            best = max(best, (total - s) * s)

        return best % (10 ** 9 + 7)

KobeSun, Dec 11, 2022

Reference

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