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    "data": {
        "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>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>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>&nbsp;</p>\n\n<p>游戏从「一号」玩家开始（「一号」玩家为红色，「二号」玩家为蓝色），最开始时，</p>\n\n<p>「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>）；</p>\n\n<p>「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>。</p>\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>&nbsp;</p>\n\n<p>之后两位玩家轮流进行操作，每一回合，玩家选择一个他之前涂好颜色的节点，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。</p>\n\n<p>如果当前玩家无法找到这样的节点来染色时，他的回合就会被跳过。</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>&nbsp;</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\n<p>&nbsp;</p>\n\n<p><strong>示例：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/08/04/1480-binary-tree-coloring-game.png\" style=\"height: 186px; width: 300px;\"></strong></p>\n\n<pre><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。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>二叉树的根节点为&nbsp;<code>root</code>，树上由 <code>n</code> 个节点，节点上的值从 <code>1</code> 到 <code>n</code> 各不相同。</li>\n\t<li><code>n</code> 为奇数。</li>\n\t<li><code>1 &lt;= x &lt;= n&nbsp;&lt;= 100</code></li>\n</ul>\n",
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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 btreeGameWinningMove(TreeNode* root, int n, int x) {\n\n    }\n};",
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                {
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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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                {
                    "lang": "Python",
                    "langSlug": "python",
                    "code": "# Definition for a binary tree node.\n# class TreeNode(object):\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution(object):\n    def btreeGameWinningMove(self, root, n, x):\n        \"\"\"\n        :type root: TreeNode\n        :type n: int\n        :type x: int\n        :rtype: bool\n        \"\"\"",
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                    "lang": "Python3",
                    "langSlug": "python3",
                    "code": "# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution:\n    def btreeGameWinningMove(self, root: Optional[TreeNode], n: int, x: int) -> bool:\n        ",
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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 *     struct TreeNode *left;\n *     struct TreeNode *right;\n * };\n */\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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                {
                    "lang": "JavaScript",
                    "langSlug": "javascript",
                    "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} n\n * @param {number} x\n * @return {boolean}\n */\nvar btreeGameWinningMove = function(root, n, x) {\n\n};",
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                {
                    "lang": "Ruby",
                    "langSlug": "ruby",
                    "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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                {
                    "lang": "Swift",
                    "langSlug": "swift",
                    "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 btreeGameWinningMove(_ root: TreeNode?, _ n: Int, _ x: Int) -> Bool {\n\n    }\n}",
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                {
                    "lang": "Go",
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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": "/**\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 btreeGameWinningMove(root: TreeNode, n: Int, x: Int): Boolean = {\n\n    }\n}",
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                {
                    "lang": "Kotlin",
                    "langSlug": "kotlin",
                    "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 btreeGameWinningMove(root: TreeNode?, n: Int, x: Int): Boolean {\n\n    }\n}",
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                {
                    "lang": "Rust",
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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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                    "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 $n\n     * @param Integer $x\n     * @return Boolean\n     */\n    function btreeGameWinningMove($root, $n, $x) {\n\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 btreeGameWinningMove(root: TreeNode | null, n: number, x: number): boolean {\n\n};",
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                    "code": "; Definition for a binary tree node.\n#|\n\n; val : integer?\n; left : (or/c tree-node? #f)\n; right : (or/c tree-node? #f)\n(struct tree-node\n  (val left right) #:mutable #:transparent)\n\n; constructor\n(define (make-tree-node [val 0])\n  (tree-node val #f #f))\n\n|#\n\n(define/contract (btree-game-winning-move root n x)\n  (-> (or/c tree-node? #f) exact-integer? exact-integer? boolean?)\n\n  )",
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                    "lang": "Erlang",
                    "langSlug": "erlang",
                    "code": "%% Definition for a binary tree node.\n%%\n%% -record(tree_node, {val = 0 :: integer(),\n%%                     left = null  :: 'null' | #tree_node{},\n%%                     right = null :: 'null' | #tree_node{}}).\n\n-spec btree_game_winning_move(Root :: #tree_node{} | null, N :: integer(), X :: integer()) -> boolean().\nbtree_game_winning_move(Root, N, X) ->\n  .",
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                    "lang": "Elixir",
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                    "code": "# Definition for a binary tree node.\n#\n# defmodule TreeNode do\n#   @type t :: %__MODULE__{\n#           val: integer,\n#           left: TreeNode.t() | nil,\n#           right: TreeNode.t() | nil\n#         }\n#   defstruct val: 0, left: nil, right: nil\n# end\n\ndefmodule Solution do\n  @spec btree_game_winning_move(root :: TreeNode.t | nil, n :: integer, x :: integer) :: boolean\n  def btree_game_winning_move(root, n, x) do\n\n  end\nend",
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            "hints": [
                "The best move y must be immediately adjacent to x, since it locks out that subtree.",
                "Can you count each of (up to) 3 different subtrees neighboring x?"
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