leetcode-problemset/leetcode-cn/originData/flip-equivalent-binary-trees.json

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        "question": {
            "questionId": "988",
            "questionFrontendId": "951",
            "categoryTitle": "Algorithms",
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            "title": "Flip Equivalent Binary Trees",
            "titleSlug": "flip-equivalent-binary-trees",
            "content": "<p>For a binary tree <strong>T</strong>, we can define a <strong>flip operation</strong> as follows: choose any node, and swap the left and right child subtrees.</p>\n\n<p>A binary tree <strong>X</strong>&nbsp;is <em>flip equivalent</em> to a binary tree <strong>Y</strong> if and only if we can make <strong>X</strong> equal to <strong>Y</strong> after some number of flip operations.</p>\n\n<p>Given the roots of two binary trees <code>root1</code> and <code>root2</code>, return <code>true</code> if the two trees are flip equivalent or <code>false</code> otherwise.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"Flipped Trees Diagram\" src=\"https://assets.leetcode.com/uploads/2018/11/29/tree_ex.png\" style=\"width: 500px; height: 220px;\" />\n<pre>\n<strong>Input:</strong> root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]\n<strong>Output:</strong> true\n<strong>Explanation: </strong>We flipped at nodes with values 1, 3, and 5.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> root1 = [], root2 = []\n<strong>Output:</strong> true\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root1 = [], root2 = [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 each tree is in the range <code>[0, 100]</code>.</li>\n\t<li>Each tree will have <strong>unique node values</strong> in the range <code>[0, 99]</code>.</li>\n</ul>\n",
            "translatedTitle": "翻转等价二叉树",
            "translatedContent": "<p>我们可以为二叉树 <strong>T</strong> 定义一个&nbsp;<strong>翻转操作&nbsp;</strong>，如下所示：选择任意节点，然后交换它的左子树和右子树。</p>\n\n<p>只要经过一定次数的翻转操作后，能使 <strong>X</strong> 等于 <strong>Y</strong>，我们就称二叉树 <strong>X</strong> <em>翻转 等价&nbsp;</em>于二叉树 <strong>Y</strong>。</p>\n\n<p>这些树由根节点&nbsp;<code>root1</code> 和 <code>root2</code>&nbsp;给出。如果两个二叉树是否是<em>翻转 等价&nbsp;</em>的函数，则返回 <code>true</code> ，否则返回 <code>false</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"Flipped Trees Diagram\" src=\"https://assets.leetcode.com/uploads/2018/11/29/tree_ex.png\" /></p>\n\n<pre>\n<strong>输入：</strong>root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]\n<strong>输出：</strong>true\n<strong>解释：</strong>我们翻转值为 1，3 以及 5 的三个节点。\n</pre>\n\n<p><strong>示例&nbsp;2:</strong></p>\n\n<pre>\n<strong>输入:</strong> root1 = [], root2 = []\n<strong>输出:</strong> true\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong> root1 = [], root2 = [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>每棵树节点数在&nbsp;<code>[0, 100]</code> 范围内</li>\n\t<li>每棵树中的每个值都是唯一的、在 <code>[0, 99]</code>&nbsp;范围内的整数</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 flipEquiv(TreeNode* root1, TreeNode* root2) {\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 flipEquiv(TreeNode root1, TreeNode root2) {\n\n    }\n}",
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                    "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 flipEquiv(self, root1, root2):\n        \"\"\"\n        :type root1: TreeNode\n        :type root2: TreeNode\n        :rtype: bool\n        \"\"\"\n        ",
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                    "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 flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:",
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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 flipEquiv(struct TreeNode* root1, struct TreeNode* root2) {\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 FlipEquiv(TreeNode root1, TreeNode root2) {\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} root1\n * @param {TreeNode} root2\n * @return {boolean}\n */\nvar flipEquiv = function(root1, root2) {\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 flipEquiv(root1: TreeNode | null, root2: TreeNode | null): boolean {\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 flipEquiv(root1: TreeNode?, root2: TreeNode?): Boolean {\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 flipEquiv(root1 *TreeNode, root2 *TreeNode) 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} root1\n# @param {TreeNode} root2\n# @return {Boolean}\ndef flip_equiv(root1, root2)\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 flip_equiv(root1: Option<Rc<RefCell<TreeNode>>>, root2: Option<Rc<RefCell<TreeNode>>>) -> bool {\n\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 (flip-equiv root1 root2)\n  (-> (or/c tree-node? #f) (or/c tree-node? #f) boolean?)\n  )",
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                    "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 flip_equiv(Root1 :: #tree_node{} | null, Root2 :: #tree_node{} | null) -> boolean().\nflip_equiv(Root1, Root2) ->\n  .",
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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 flip_equiv(root1 :: TreeNode.t | nil, root2 :: TreeNode.t | nil) :: boolean\n  def flip_equiv(root1, root2) do\n    \n  end\nend",
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