leetcode-problemset/leetcode-cn/originData/check-balance-lcci.json

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        "question": {
            "questionId": "100176",
            "questionFrontendId": "面试题 04.04",
            "categoryTitle": "LCCI",
            "boundTopicId": 46211,
            "title": "Check Balance LCCI",
            "titleSlug": "check-balance-lcci",
            "content": "<p>Implement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one.</p>\r\n\r\n<p><br />\r\n<strong>Example 1:</strong></p>\r\n\r\n<pre>\r\nGiven tree [3,9,20,null,null,15,7]\r\n    3\r\n   / \\\r\n  9  20\r\n    /  \\\r\n   15   7\r\nreturn true.</pre>\r\n\r\n<p><strong>Example 2:</strong></p>\r\n\r\n<pre>\r\nGiven [1,2,2,3,3,null,null,4,4]\r\n      1\r\n     / \\\r\n    2   2\r\n   / \\\r\n  3   3\r\n / \\\r\n4   4\r\nreturn&nbsp;false.</pre>\r\n\r\n<p>&nbsp;</p>\r\n",
            "translatedTitle": "检查平衡性",
            "translatedContent": "<p>实现一个函数，检查二叉树是否平衡。在这个问题中，平衡树的定义如下：任意一个节点，其两棵子树的高度差不超过 1。</p><br><strong>示例 1:</strong><pre>给定二叉树 [3,9,20,null,null,15,7]<br>    3<br>   / &#92<br>  9  20<br>    /  &#92<br>   15   7<br>返回 true 。</pre><strong>示例 2:</strong><br><pre>给定二叉树 [1,2,2,3,3,null,null,4,4]<br>      1<br>     / &#92<br>    2   2<br>   / &#92<br>  3   3<br> / &#92<br>4   4<br>返回 false 。</pre>",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode(object):\n#     def __init__(self, x):\n#         self.val = x\n#         self.left = None\n#         self.right = None\n\nclass Solution(object):\n    def isBalanced(self, root):\n        \"\"\"\n        :type root: 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 */\n\n\nbool isBalanced(struct 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 x) { val = x; }\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) {\n *     this.val = val;\n *     this.left = this.right = null;\n * }\n */\n/**\n * @param {TreeNode} root\n * @return {boolean}\n */\nvar isBalanced = function(root) {\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": "/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n *     Val int\n *     Left *TreeNode\n *     Right *TreeNode\n * }\n */\nfunc isBalanced(root *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)\n#         @val = val\n#         @left, @right = nil, nil\n#     end\n# end\n\n# @param {TreeNode} root\n# @return {Boolean}\ndef is_balanced(root)\n\nend",
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                    "code": "/**\n * Definition for a binary tree node.\n * class TreeNode(var _value: Int) {\n *   var value: Int = _value\n *   var left: TreeNode = null\n *   var right: TreeNode = null\n * }\n */\nobject Solution {\n    def isBalanced(root: TreeNode): Boolean = {\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 is_balanced(root: 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 (is-balanced root)\n  (-> (or/c tree-node? #f) boolean?)\n\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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            "hints": [
                "考虑平衡树的定义。你可以检查单个节点的条件吗？你可以检查每个节点吗？",
                "如果你开发了一个蛮力解法，请注意它的运行时间。如果你是用于计算每个节点的子树的高度，那么该算法会很低效。",
                "如果你可以修改二叉树节点类，允许节点存储子树的高度，会如何？",
                "你不需要修改二叉树类来存储子树的高度。递归函数是否可以计算每个子树的高度，同时检查节点是否平衡？尝试让函数返回多个值。",
                "其实，你只需要一个checkHeight函数即可，它既可以计算高度，也可以平衡检查。可以使用整数返回值表示两者。"
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