mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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189 lines
26 KiB
JSON
189 lines
26 KiB
JSON
{
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"questionId": "1475",
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"questionFrontendId": "1373",
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"categoryTitle": "Algorithms",
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"boundTopicId": 131576,
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"title": "Maximum Sum BST in Binary Tree",
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"titleSlug": "maximum-sum-bst-in-binary-tree",
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"content": "<p>Given a <strong>binary tree</strong> <code>root</code>, return <em>the maximum sum of all keys of <strong>any</strong> sub-tree which is also a Binary Search Tree (BST)</em>.</p>\n\n<p>Assume a BST is defined as follows:</p>\n\n<ul>\n\t<li>The left subtree of a node contains only nodes with keys <strong>less than</strong> the node's key.</li>\n\t<li>The right subtree of a node contains only nodes with keys <strong>greater than</strong> the node's key.</li>\n\t<li>Both the left and right subtrees must also be binary search trees.</li>\n</ul>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/30/sample_1_1709.png\" style=\"width: 320px; height: 250px;\" /></p>\n\n<pre>\n<strong>Input:</strong> root = [1,4,3,2,4,2,5,null,null,null,null,null,null,4,6]\n<strong>Output:</strong> 20\n<strong>Explanation:</strong> Maximum sum in a valid Binary search tree is obtained in root node with key equal to 3.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/30/sample_2_1709.png\" style=\"width: 134px; height: 180px;\" /></p>\n\n<pre>\n<strong>Input:</strong> root = [4,3,null,1,2]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> Maximum sum in a valid Binary search tree is obtained in a single root node with key equal to 2.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [-4,-2,-5]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> All values are negatives. Return an empty BST.\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>[1, 4 * 10<sup>4</sup>]</code>.</li>\n\t<li><code>-4 * 10<sup>4</sup> <= Node.val <= 4 * 10<sup>4</sup></code></li>\n</ul>\n",
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"translatedTitle": "二叉搜索子树的最大键值和",
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"translatedContent": "<p>给你一棵以 <code>root</code> 为根的 <strong>二叉树</strong> ，请你返回 <strong>任意</strong> 二叉搜索子树的最大键值和。</p>\n\n<p>二叉搜索树的定义如下：</p>\n\n<ul>\n\t<li>任意节点的左子树中的键值都 <strong>小于</strong> 此节点的键值。</li>\n\t<li>任意节点的右子树中的键值都 <strong>大于</strong> 此节点的键值。</li>\n\t<li>任意节点的左子树和右子树都是二叉搜索树。</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/03/07/sample_1_1709.png\" style=\"height: 250px; width: 320px;\" /></p>\n\n<pre>\n<strong>输入：</strong>root = [1,4,3,2,4,2,5,null,null,null,null,null,null,4,6]\n<strong>输出：</strong>20\n<strong>解释：</strong>键值为 3 的子树是和最大的二叉搜索树。\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/2020/03/07/sample_2_1709.png\" style=\"height: 180px; width: 134px;\" /></p>\n\n<pre>\n<strong>输入：</strong>root = [4,3,null,1,2]\n<strong>输出：</strong>2\n<strong>解释：</strong>键值为 2 的单节点子树是和最大的二叉搜索树。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [-4,-2,-5]\n<strong>输出：</strong>0\n<strong>解释：</strong>所有节点键值都为负数，和最大的二叉搜索树为空。\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [2,1,3]\n<strong>输出：</strong>6\n</pre>\n\n<p><strong>示例 5：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [5,4,8,3,null,6,3]\n<strong>输出：</strong>7\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>每棵树有 <code>1</code> 到 <code>40000</code> 个节点。</li>\n\t<li>每个节点的键值在 <code>[-4 * 10^4 , 4 * 10^4]</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 int val;\n * public TreeNode left;\n * public TreeNode right;\n * public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {\n * this.val = val;\n * this.left = left;\n * this.right = right;\n * }\n * }\n */\npublic class Solution {\n public int MaxSumBST(TreeNode root) {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * function TreeNode(val, left, right) {\n * this.val = (val===undefined ? 0 : val)\n * this.left = (left===undefined ? null : left)\n * this.right = (right===undefined ? null : right)\n * }\n */\n/**\n * @param {TreeNode} root\n * @return {number}\n */\nvar maxSumBST = function(root) {\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_sum_bst(root)\n\nend",
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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 maxSumBST(_ root: TreeNode?) -> Int {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n * Val int\n * Left *TreeNode\n * Right *TreeNode\n * }\n */\nfunc maxSumBST(root *TreeNode) int {\n\n}",
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"code": "/**\n * Example:\n * var ti = TreeNode(5)\n * var v = ti.`val`\n * Definition for a binary tree node.\n * class TreeNode(var `val`: Int) {\n * var left: TreeNode? = null\n * var right: TreeNode? = null\n * }\n */\nclass Solution {\n fun maxSumBST(root: TreeNode?): Int {\n\n }\n}",
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"code": "// Definition for a binary tree node.\n// #[derive(Debug, PartialEq, Eq)]\n// pub struct TreeNode {\n// pub val: i32,\n// pub left: Option<Rc<RefCell<TreeNode>>>,\n// pub right: Option<Rc<RefCell<TreeNode>>>,\n// }\n//\n// impl TreeNode {\n// #[inline]\n// pub fn new(val: i32) -> Self {\n// TreeNode {\n// val,\n// left: None,\n// right: None\n// }\n// }\n// }\nuse std::rc::Rc;\nuse std::cell::RefCell;\nimpl Solution {\n pub fn max_sum_bst(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {\n\n }\n}",
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"Create a datastructure with 4 parameters: (sum, isBST, maxLeft, minLeft).",
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"In each node compute theses parameters, following the conditions of a Binary Search Tree."
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