mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-03-15 00:32:24 +08:00
198 lines
26 KiB
JSON
198 lines
26 KiB
JSON
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{
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"data": {
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"question": {
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"questionId": "653",
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"questionFrontendId": "653",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1777,
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"title": "Two Sum IV - Input is a BST",
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"titleSlug": "two-sum-iv-input-is-a-bst",
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"content": "<p>Given the <code>root</code> of a Binary Search Tree and a target number <code>k</code>, return <em><code>true</code> if there exist two elements in the BST such that their sum is equal to the given target</em>.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/09/21/sum_tree_1.jpg\" style=\"width: 400px; height: 229px;\" />\n<pre>\n<strong>Input:</strong> root = [5,3,6,2,4,null,7], k = 9\n<strong>Output:</strong> true\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/09/21/sum_tree_2.jpg\" style=\"width: 400px; height: 229px;\" />\n<pre>\n<strong>Input:</strong> root = [5,3,6,2,4,null,7], k = 28\n<strong>Output:</strong> false\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, 10<sup>4</sup>]</code>.</li>\n\t<li><code>-10<sup>4</sup> <= Node.val <= 10<sup>4</sup></code></li>\n\t<li><code>root</code> is guaranteed to be a <strong>valid</strong> binary search tree.</li>\n\t<li><code>-10<sup>5</sup> <= k <= 10<sup>5</sup></code></li>\n</ul>\n",
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"translatedTitle": "两数之和 IV - 输入 BST",
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"translatedContent": "<p>给定一个二叉搜索树 <code>root</code> 和一个目标结果 <code>k</code>，如果 BST 中存在两个元素且它们的和等于给定的目标结果，则返回 <code>true</code>。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/09/21/sum_tree_1.jpg\" style=\"height: 229px; width: 400px;\" />\n<pre>\n<strong>输入:</strong> root = [5,3,6,2,4,null,7], k = 9\n<strong>输出:</strong> true\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/09/21/sum_tree_2.jpg\" style=\"height: 229px; width: 400px;\" />\n<pre>\n<strong>输入:</strong> root = [5,3,6,2,4,null,7], k = 28\n<strong>输出:</strong> false\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>二叉树的节点个数的范围是 <code>[1, 10<sup>4</sup>]</code>.</li>\n\t<li><code>-10<sup>4</sup> <= Node.val <= 10<sup>4</sup></code></li>\n\t<li><code>root</code> 为二叉搜索树</li>\n\t<li><code>-10<sup>5</sup> <= k <= 10<sup>5</sup></code></li>\n</ul>\n",
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"difficulty": "Easy",
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"similarQuestions": "[{\"title\": \"Two Sum\", \"titleSlug\": \"two-sum\", \"difficulty\": \"Easy\", \"translatedTitle\": \"\\u4e24\\u6570\\u4e4b\\u548c\"}, {\"title\": \"Two Sum II - Input Array Is Sorted\", \"titleSlug\": \"two-sum-ii-input-array-is-sorted\", \"difficulty\": \"Medium\", \"translatedTitle\": \"\\u4e24\\u6570\\u4e4b\\u548c II - \\u8f93\\u5165\\u6709\\u5e8f\\u6570\\u7ec4\"}, {\"title\": \"Two Sum III - Data structure design\", \"titleSlug\": \"two-sum-iii-data-structure-design\", \"difficulty\": \"Easy\", \"translatedTitle\": \"\\u4e24\\u6570\\u4e4b\\u548c III - \\u6570\\u636e\\u7ed3\\u6784\\u8bbe\\u8ba1\"}, {\"title\": \"Two Sum BSTs\", \"titleSlug\": \"two-sum-bsts\", \"difficulty\": \"Medium\", \"translatedTitle\": \"\\u67e5\\u627e\\u4e24\\u68f5\\u4e8c\\u53c9\\u641c\\u7d22\\u6811\\u4e4b\\u548c\"}]",
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"topicTags": [
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"name": "Tree",
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"translatedName": "树",
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"name": "Depth-First Search",
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"slug": "depth-first-search",
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"translatedName": "深度优先搜索",
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"name": "Breadth-First Search",
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"slug": "breadth-first-search",
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"translatedName": "广度优先搜索",
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"name": "Binary Search Tree",
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"slug": "binary-search-tree",
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"translatedName": "二叉搜索树",
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"name": "Hash Table",
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"slug": "hash-table",
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"translatedName": "哈希表",
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"__typename": "TopicTagNode"
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"name": "Two Pointers",
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"slug": "two-pointers",
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"translatedName": "双指针",
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"name": "Binary Tree",
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"slug": "binary-tree",
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"translatedName": "二叉树",
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"lang": "C++",
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"code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * TreeNode *left;\n * TreeNode *right;\n * TreeNode() : val(0), left(nullptr), right(nullptr) {}\n * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\npublic:\n bool findTarget(TreeNode* root, int k) {\n\n }\n};",
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"code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * struct TreeNode *left;\n * struct TreeNode *right;\n * };\n */\n\n\nbool findTarget(struct TreeNode* root, int k){\n\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 bool FindTarget(TreeNode root, int k) {\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 * @param {number} k\n * @return {boolean}\n */\nvar findTarget = function(root, k) {\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# @param {Integer} k\n# @return {Boolean}\ndef find_target(root, k)\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 findTarget(_ root: TreeNode?, _ k: Int) -> Bool {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {\n * var value: Int = _value\n * var left: TreeNode = _left\n * var right: TreeNode = _right\n * }\n */\nobject Solution {\n def findTarget(root: TreeNode, k: Int): Boolean = {\n\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 findTarget(root: TreeNode?, k: Int): Boolean {\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 find_target(root: Option<Rc<RefCell<TreeNode>>>, k: i32) -> bool {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * public $val = null;\n * public $left = null;\n * public $right = null;\n * function __construct($val = 0, $left = null, $right = null) {\n * $this->val = $val;\n * $this->left = $left;\n * $this->right = $right;\n * }\n * }\n */\nclass Solution {\n\n /**\n * @param TreeNode $root\n * @param Integer $k\n * @return Boolean\n */\n function findTarget($root, $k) {\n\n }\n}",
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