10000 LeetCode/109. Balanced Binary Tree at master · algorithms-forks/LeetCode · GitHub
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< 7377 script type="application/json" data-target="react-app.embeddedData">{"payload":{"allShortcutsEnabled":false,"path":"109. Balanced Binary Tree","repo":{"id":490101071,"defaultBranch":"master","name":"LeetCode","ownerLogin":"algorithms-forks","currentUserCanPush":false,"isFork":true,"isEmpty":false,"createdAt":"2022-05-09T01:47:36.000Z","ownerAvatar":"https://avatars.githubusercontent.com/u/105182575?v=4","public":true,"private":false,"isOrgOwned":true},"currentUser":null,"refInfo":{"name":"master","listCacheKey":"v0:1652060858.178429","canEdit":false,"refType":"branch","currentOid":"1756256d7e619164076bbf358c8f7ca68cd8bd79"},"tree":{"items":[{"name":"README.md","path":"109. Balanced Binary Tree/README.md","contentType":"file"},{"name":"main.cc","path":"109. Balanced Binary Tree/main.cc","contentType":"file"},{"name":"solution.h","path":"109. Balanced Binary Tree/solution.h","contentType":"file"}],"templateDirectorySuggestionUrl":null,"readme":{"displayName":"README.md","richText":"\u003carticle class=\"markdown-body entry-content container-lg\" itemprop=\"text\"\u003e\u003cp dir=\"auto\"\u003e这道题，我昨天就在想了。毕竟平衡搜索树只不过是特殊的平衡二叉树罢了。是否平衡的关键在于题目里给出的这个定义：\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp dir=\"auto\"\u003ea height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003cp dir=\"auto\"\u003e所以这个问题很重要的一个子问题就是：如何求二叉树的高度。假设我们找到了高度的方法，那么核心判断就是：\u003c/p\u003e\n\u003cdiv class=\"highlight highlight-source-c++ notranslate position-relative overflow-auto\" dir=\"auto\" data-snippet-clipboard-copy-content=\"std::abs(height(root-\u0026gt;left), height(root-\u0026gt;right)) \u0026lt;= 1\"\u003e\u003cpre\u003e\u003cspan class=\"pl-en\"\u003estd::abs\u003c/span\u003e(height(root-\u0026gt;left), height(root-\u0026gt;right)) \u0026lt;= 1\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e所以我几乎是瞬间写出了\u003ca href=\"/algorithms-forks/LeetCode/blob/e6906ef932bda154ad17518e6d097c904045048f/22.%20Balanced%20Binary%20Tree/solution.h#L22-L32\"\u003e解决方法\u003c/a\u003e（递归真的是人类自然而然的思路呀）。\n但往往自然的思路，有时会伴随着效率低下，虽然被 LeetCode 认可了的（可AC）。\u003c/p\u003e\n\u003chr\u003e\n\u003cp dir=\"auto\"\u003e于是我尝试缩减递归次数，能想到的办法就是及早的判断出非平衡，然后跳出递归。那么又回到上面的\u003cstrong\u003e判断核心\u003c/strong\u003e上，只要那个条件不成立，我就应该跳出递归。可是如何能跳出递归呢？\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e我能想到的方法就是设一个 bool 成员变量，然后一旦为 false， 立刻停止。这就是目前我的方案了。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e希望抛砖引玉，了解更多的方案。\u003c/p\u003e\n\u003chr\u003e\n\u003cp dir=\"auto\"\u003e最近持续不断的做有关 \u003cstrong\u003etree\u003c/strong\u003e 的题目，让我深深感觉到，在树算法中，不用递归简直是找死，自建栈也好，自建队列也好，\n都要维护一大堆东西，在效率上不仅没有显著的提升，反而让程序晦涩难懂，表意不明。如何抉择呢？我一定还是偏向于使用递归的。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e在效率没有明显提升的情况下，递归表达意图的能力，实在是让人无法割舍。\u003c/p\u003e\n\u003c/article\u003e","errorMessage":null,"headerInfo":{"toc":[],"siteNavLoginPath":"/login?return_to=https%3A%2F%2Fgithub.com%2Falgorithms-forks%2FLeetCode%2Ftree%2Fmaster%2F109.%2520Balanced%2520Binary%2520Tree"}},"totalCount":3,"showBranchInfobar":true},"fileTree":{"":{"items":[{"name":"000. 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