mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-26 02:00:27 +08:00
189 lines
28 KiB
JSON
189 lines
28 KiB
JSON
{
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"questionId": "2461",
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"questionFrontendId": "2385",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1759751,
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"title": "Amount of Time for Binary Tree to Be Infected",
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"titleSlug": "amount-of-time-for-binary-tree-to-be-infected",
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"content": "<p>You are given the <code>root</code> of a binary tree with <strong>unique</strong> values, and an integer <code>start</code>. At minute <code>0</code>, an <strong>infection</strong> starts from the node with value <code>start</code>.</p>\n\n<p>Each minute, a node becomes infected if:</p>\n\n<ul>\n\t<li>The node is currently uninfected.</li>\n\t<li>The node is adjacent to an infected node.</li>\n</ul>\n\n<p>Return <em>the number of minutes needed for the entire tree to be infected.</em></p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/06/25/image-20220625231744-1.png\" style=\"width: 400px; height: 306px;\" />\n<pre>\n<strong>Input:</strong> root = [1,5,3,null,4,10,6,9,2], start = 3\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> The following nodes are infected during:\n- Minute 0: Node 3\n- Minute 1: Nodes 1, 10 and 6\n- Minute 2: Node 5\n- Minute 3: Node 4\n- Minute 4: Nodes 9 and 2\nIt takes 4 minutes for the whole tree to be infected so we return 4.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/06/25/image-20220625231812-2.png\" style=\"width: 75px; height: 66px;\" />\n<pre>\n<strong>Input:</strong> root = [1], start = 1\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> At minute 0, the only node in the tree is infected so we return 0.\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>5</sup>]</code>.</li>\n\t<li><code>1 <= Node.val <= 10<sup>5</sup></code></li>\n\t<li>Each node has a <strong>unique</strong> value.</li>\n\t<li>A node with a value of <code>start</code> exists in the tree.</li>\n</ul>\n",
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"translatedTitle": "感染二叉树需要的总时间",
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"translatedContent": "<p>给你一棵二叉树的根节点 <code>root</code> ，二叉树中节点的值 <strong>互不相同</strong> 。另给你一个整数 <code>start</code> 。在第 <code>0</code> 分钟，<strong>感染</strong> 将会从值为 <code>start</code> 的节点开始爆发。</p>\n\n<p>每分钟，如果节点满足以下全部条件，就会被感染：</p>\n\n<ul>\n\t<li>节点此前还没有感染。</li>\n\t<li>节点与一个已感染节点相邻。</li>\n</ul>\n\n<p>返回感染整棵树需要的分钟数<em>。</em></p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/06/25/image-20220625231744-1.png\" style=\"width: 400px; height: 306px;\">\n<pre><strong>输入：</strong>root = [1,5,3,null,4,10,6,9,2], start = 3\n<strong>输出：</strong>4\n<strong>解释：</strong>节点按以下过程被感染：\n- 第 0 分钟：节点 3\n- 第 1 分钟：节点 1、10、6\n- 第 2 分钟：节点5\n- 第 3 分钟：节点 4\n- 第 4 分钟：节点 9 和 2\n感染整棵树需要 4 分钟，所以返回 4 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/06/25/image-20220625231812-2.png\" style=\"width: 75px; height: 66px;\">\n<pre><strong>输入：</strong>root = [1], start = 1\n<strong>输出：</strong>0\n<strong>解释：</strong>第 0 分钟，树中唯一一个节点处于感染状态，返回 0 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中节点的数目在范围 <code>[1, 10<sup>5</sup>]</code> 内</li>\n\t<li><code>1 <= Node.val <= 10<sup>5</sup></code></li>\n\t<li>每个节点的值 <strong>互不相同</strong></li>\n\t<li>树中必定存在值为 <code>start</code> 的节点</li>\n</ul>\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} start\n * @return {number}\n */\nvar amountOfTime = function(root, start) {\n\n};",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * val: number\n * left: TreeNode | null\n * right: TreeNode | null\n * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {\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\nfunction amountOfTime(root: TreeNode | null, start: number): number {\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 $start\n * @return Integer\n */\n function amountOfTime($root, $start) {\n\n }\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 amountOfTime(_ root: TreeNode?, _ start: Int) -> Int {\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 amountOfTime(root: TreeNode?, start: Int): Int {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * int val;\n * TreeNode? left;\n * TreeNode? right;\n * TreeNode([this.val = 0, this.left, this.right]);\n * }\n */\nclass Solution {\n int amountOfTime(TreeNode? root, int start) {\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# @param {Integer} start\n# @return {Integer}\ndef amount_of_time(root, start)\n\nend",
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"Convert the tree to an undirected graph to make it easier to handle.",
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"Use BFS starting at the start node to find the distance between each node and the start node. The answer is the maximum distance."
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