leetcode-problemset/leetcode-cn/originData/ping-heng-er-cha-shu-lcof.json

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        "question": {
            "questionId": "100342",
            "questionFrontendId": "LCR 176",
            "categoryTitle": "Algorithms",
            "boundTopicId": 88134,
            "title": "判断是否为平衡二叉树",
            "titleSlug": "ping-heng-er-cha-shu-lcof",
            "content": "<p>English description is not available for the problem. Please switch to Chinese.</p>\r\n",
            "translatedTitle": "判断是否为平衡二叉树",
            "translatedContent": "<p>输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [3,9,20,null,null,15,7]\n<strong>输出：</strong>true \n<strong>解释：</strong>如下图\n</pre>\n\n<p><img alt=\"\" src=\"https://pic.leetcode.cn/1695102431-vbmWJn-image.png\" style=\"height: 281px; width: 500px;\" /><br />\n<br />\n<strong>示例 2:</strong></p>\n\n<pre>\n输入：root = [1,2,2,3,3,null,null,4,4]\n输出：false\n解释：如下图\n</pre>\n<img alt=\"\" src=\"https://pic.leetcode.cn/1695102434-WlaxCo-image.png\" style=\"height: 281px; width: 500px;\" />\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= 树的结点个数 &lt;= 10000</code></li>\n</ul>\n\n<p>注意：本题与主站 110&nbsp;题相同：<a href=\"https://leetcode-cn.com/problems/balanced-binary-tree/\">https://leetcode-cn.com/problems/balanced-binary-tree/</a></p>\n",
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                    "translatedName": "树",
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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 isBalanced(TreeNode* root) {\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 isBalanced(self, root):\n        \"\"\"\n        :type root: Optional[TreeNode]\n        :rtype: bool\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 isBalanced(struct TreeNode* root) {\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 IsBalanced(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 {boolean}\n */\nvar isBalanced = function(root) {\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 isBalanced(root: 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 isBalanced(root: TreeNode?): Boolean {\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# @return {Boolean}\ndef is_balanced(root)\n\nend",
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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 is_balanced(Root :: #tree_node{} | null) -> boolean().\nis_balanced(Root) ->\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 is_balanced(root :: TreeNode.t | nil) :: boolean\n  def is_balanced(root) do\n    \n  end\nend",
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