leetcode-problemset/leetcode-cn/originData/binary-tree-coloring-game.json

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        "question": {
            "questionId": "1248",
            "questionFrontendId": "1145",
            "categoryTitle": "Algorithms",
            "boundTopicId": 18046,
            "title": "Binary Tree Coloring Game",
            "titleSlug": "binary-tree-coloring-game",
            "content": "<p>Two players play a turn based game on a binary tree. We are given the <code>root</code> of this binary tree, and the number of nodes <code>n</code> in the tree. <code>n</code> is odd, and each node has a distinct value from <code>1</code> to <code>n</code>.</p>\n\n<p>Initially, the first player names a value <code>x</code> with <code>1 &lt;= x &lt;= n</code>, and the second player names a value <code>y</code> with <code>1 &lt;= y &lt;= n</code> and <code>y != x</code>. The first player colors the node with value <code>x</code> red, and the second player colors the node with value <code>y</code> blue.</p>\n\n<p>Then, the players take turns starting with the first player. In each turn, that player chooses a node of their color (red if player 1, blue if player 2) and colors an <strong>uncolored</strong> neighbor of the chosen node (either the left child, right child, or parent of the chosen node.)</p>\n\n<p>If (and only if) a player cannot choose such a node in this way, they must pass their turn. If both players pass their turn, the game ends, and the winner is the player that colored more nodes.</p>\n\n<p>You are the second player. If it is possible to choose such a <code>y</code> to ensure you win the game, return <code>true</code>. If it is not possible, return <code>false</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/08/01/1480-binary-tree-coloring-game.png\" style=\"width: 500px; height: 310px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3\n<strong>Output:</strong> true\n<strong>Explanation: </strong>The second player can choose the node with value 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [1,2,3], n = 3, x = 1\n<strong>Output:</strong> false\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is <code>n</code>.</li>\n\t<li><code>1 &lt;= x &lt;= n &lt;= 100</code></li>\n\t<li><code>n</code> is odd.</li>\n\t<li>1 &lt;= Node.val &lt;= n</li>\n\t<li>All the values of the tree are <strong>unique</strong>.</li>\n</ul>\n",
            "translatedTitle": "二叉树着色游戏",
            "translatedContent": "<p>有两位极客玩家参与了一场「二叉树着色」的游戏。游戏中，给出二叉树的根节点&nbsp;<code>root</code>，树上总共有 <code>n</code> 个节点，且 <code>n</code> 为奇数，其中每个节点上的值从&nbsp;<code>1</code> 到&nbsp;<code>n</code>&nbsp;各不相同。</p>\n\n<p>最开始时：</p>\n\n<ul>\n\t<li>「一号」玩家从 <code>[1, n]</code>&nbsp;中取一个值&nbsp;<code>x</code>（<code>1 &lt;= x &lt;= n</code>）；</li>\n\t<li>「二号」玩家也从&nbsp;<code>[1, n]</code>&nbsp;中取一个值&nbsp;<code>y</code>（<code>1 &lt;= y &lt;= n</code>）且&nbsp;<code>y != x</code>。</li>\n</ul>\n\n<p>「一号」玩家给值为&nbsp;<code>x</code>&nbsp;的节点染上红色，而「二号」玩家给值为&nbsp;<code>y</code>&nbsp;的节点染上蓝色。</p>\n\n<p>之后两位玩家轮流进行操作，「一号」玩家先手。每一回合，玩家选择一个被他染过色的节点，将所选节点一个 <strong>未着色 </strong>的邻节点（即左右子节点、或父节点）进行染色（「一号」玩家染红色，「二号」玩家染蓝色）。</p>\n\n<p>如果（且仅在此种情况下）当前玩家无法找到这样的节点来染色时，其回合就会被跳过。</p>\n\n<p>若两个玩家都没有可以染色的节点时，游戏结束。着色节点最多的那位玩家获得胜利 ✌️。</p>\n\n<p>现在，假设你是「二号」玩家，根据所给出的输入，假如存在一个&nbsp;<code>y</code>&nbsp;值可以确保你赢得这场游戏，则返回&nbsp;<code>true</code> ；若无法获胜，就请返回 <code>false</code> 。</p>\n&nbsp;\n\n<p><strong class=\"example\">示例 1 ：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/08/01/1480-binary-tree-coloring-game.png\" style=\"width: 500px; height: 310px;\" />\n<pre>\n<strong>输入：</strong>root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3\n<strong>输出：</strong>true\n<strong>解释：</strong>第二个玩家可以选择值为 2 的节点。</pre>\n\n<p><strong class=\"example\">示例 2 ：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [1,2,3], n = 3, x = 1\n<strong>输出：</strong>false\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中节点数目为 <code>n</code></li>\n\t<li><code>1 &lt;= x &lt;= n &lt;= 100</code></li>\n\t<li><code>n</code> 是奇数</li>\n\t<li><code>1 &lt;= Node.val &lt;= n</code></li>\n\t<li>树中所有值 <strong>互不相同</strong></li>\n</ul>\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    bool btreeGameWinningMove(TreeNode* root, int n, int x) {\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 boolean btreeGameWinningMove(TreeNode root, int n, int x) {\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 */\nbool btreeGameWinningMove(struct TreeNode* root, int n, int x) {\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 BtreeGameWinningMove(TreeNode root, int n, int x) {\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 btreeGameWinningMove(root *TreeNode, n int, x int) bool {\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} n\n# @param {Integer} x\n# @return {Boolean}\ndef btree_game_winning_move(root, n, x)\n\nend",
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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 btree_game_winning_move(root: Option<Rc<RefCell<TreeNode>>>, n: i32, x: i32) -> bool {\n\n    }\n}",
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                "The best move y must be immediately adjacent to x, since it locks out that subtree.",
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