mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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184 lines
25 KiB
JSON
184 lines
25 KiB
JSON
{
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"questionId": "1008",
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"questionFrontendId": "968",
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"categoryTitle": "Algorithms",
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"boundTopicId": 2735,
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"title": "Binary Tree Cameras",
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"titleSlug": "binary-tree-cameras",
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"content": "<p>You are given the <code>root</code> of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.</p>\n\n<p>Return <em>the minimum number of cameras needed to monitor all nodes of the tree</em>.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png\" style=\"width: 138px; height: 163px;\" />\n<pre>\n<strong>Input:</strong> root = [0,0,null,0,0]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> One camera is enough to monitor all nodes if placed as shown.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png\" style=\"width: 139px; height: 312px;\" />\n<pre>\n<strong>Input:</strong> root = [0,0,null,0,null,0,null,null,0]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.\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, 1000]</code>.</li>\n\t<li><code>Node.val == 0</code></li>\n</ul>\n",
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"translatedTitle": "监控二叉树",
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"translatedContent": "<p>给定一个二叉树，我们在树的节点上安装摄像头。</p>\n\n<p>节点上的每个摄影头都可以监视<strong>其父对象、自身及其直接子对象。</strong></p>\n\n<p>计算监控树的所有节点所需的最小摄像头数量。</p>\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/2018/12/29/bst_cameras_01.png\" style=\"height: 163px; width: 138px;\"></p>\n\n<pre><strong>输入：</strong>[0,0,null,0,0]\n<strong>输出：</strong>1\n<strong>解释：</strong>如图所示，一台摄像头足以监控所有节点。\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/2018/12/29/bst_cameras_02.png\" style=\"height: 312px; width: 139px;\"></p>\n\n<pre><strong>输入：</strong>[0,0,null,0,null,0,null,null,0]\n<strong>输出：</strong>2\n<strong>解释：</strong>需要至少两个摄像头来监视树的所有节点。 上图显示了摄像头放置的有效位置之一。\n</pre>\n\n<p><br>\n<strong>提示：</strong></p>\n\n<ol>\n\t<li>给定树的节点数的范围是 <code>[1, 1000]</code>。</li>\n\t<li>每个节点的值都是 0。</li>\n</ol>\n",
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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 int minCameraCover(TreeNode* root) {\n\n }\n};",
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"code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * int val;\n * TreeNode left;\n * TreeNode right;\n * TreeNode() {}\n * TreeNode(int val) { this.val = val; }\n * TreeNode(int val, TreeNode left, TreeNode right) {\n * this.val = val;\n * this.left = left;\n * this.right = right;\n * }\n * }\n */\nclass Solution {\n public int minCameraCover(TreeNode root) {\n\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 int MinCameraCover(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 minCameraCover = 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 min_camera_cover(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 minCameraCover(_ 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 minCameraCover(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 minCameraCover(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 min_camera_cover(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {\n\n }\n}",
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