mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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182 lines
27 KiB
JSON
182 lines
27 KiB
JSON
{
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"questionId": "1465",
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"questionFrontendId": "1339",
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"categoryTitle": "Algorithms",
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"boundTopicId": 81670,
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"title": "Maximum Product of Splitted Binary Tree",
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"titleSlug": "maximum-product-of-splitted-binary-tree",
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"content": "<p>Given the <code>root</code> 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.</p>\n\n<p>Return <em>the maximum product of the sums of the two subtrees</em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p><strong>Note</strong> that you need to maximize the answer before taking the mod and not after taking it.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/21/sample_1_1699.png\" style=\"width: 500px; height: 167px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3,4,5,6]\n<strong>Output:</strong> 110\n<strong>Explanation:</strong> Remove the red edge and get 2 binary trees with sum 11 and 10. Their product is 110 (11*10)\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/21/sample_2_1699.png\" style=\"width: 500px; height: 211px;\" />\n<pre>\n<strong>Input:</strong> root = [1,null,2,3,4,null,null,5,6]\n<strong>Output:</strong> 90\n<strong>Explanation:</strong> Remove the red edge and get 2 binary trees with sum 15 and 6.Their product is 90 (15*6)\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is in the range <code>[2, 5 * 10<sup>4</sup>]</code>.</li>\n\t<li><code>1 <= Node.val <= 10<sup>4</sup></code></li>\n</ul>\n",
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"translatedTitle": "分裂二叉树的最大乘积",
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"translatedContent": "<p>给你一棵二叉树，它的根为 <code>root</code> 。请你删除 1 条边，使二叉树分裂成两棵子树，且它们子树和的乘积尽可能大。</p>\n\n<p>由于答案可能会很大，请你将结果对 10^9 + 7 取模后再返回。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/02/02/sample_1_1699.png\" style=\"height: 200px; width: 495px;\"></strong></p>\n\n<pre><strong>输入：</strong>root = [1,2,3,4,5,6]\n<strong>输出：</strong>110\n<strong>解释：</strong>删除红色的边，得到 2 棵子树，和分别为 11 和 10 。它们的乘积是 110 （11*10）\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/02/02/sample_2_1699.png\" style=\"height: 200px; width: 495px;\"></p>\n\n<pre><strong>输入：</strong>root = [1,null,2,3,4,null,null,5,6]\n<strong>输出：</strong>90\n<strong>解释：</strong>移除红色的边，得到 2 棵子树，和分别是 15 和 6 。它们的乘积为 90 （15*6）\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>root = [2,3,9,10,7,8,6,5,4,11,1]\n<strong>输出：</strong>1025\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre><strong>输入：</strong>root = [1,1]\n<strong>输出：</strong>1\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>每棵树最多有 <code>50000</code> 个节点，且至少有 <code>2</code> 个节点。</li>\n\t<li>每个节点的值在 <code>[1, 10000]</code> 之间。</li>\n</ul>\n",
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"code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * public var val: Int\n * public var left: TreeNode?\n * public var right: TreeNode?\n * public init() { self.val = 0; self.left = nil; self.right = nil; }\n * public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }\n * public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {\n * self.val = val\n * self.left = left\n * self.right = right\n * }\n * }\n */\nclass Solution {\n func maxProduct(_ root: TreeNode?) -> Int {\n\n }\n}",
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"code": "# Definition for a binary tree node.\n# class TreeNode\n# attr_accessor :val, :left, :right\n# def initialize(val = 0, left = nil, right = nil)\n# @val = val\n# @left = left\n# @right = right\n# end\n# end\n# @param {TreeNode} root\n# @return {Integer}\ndef max_product(root)\n\nend",
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"If we know the sum of a subtree, the answer is max( (total_sum - subtree_sum) * subtree_sum) in each node."
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