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{"payload":{"allShortcutsEnabled":false,"path":"098. Recover Binary Search Tree","repo":{"id":198780537,"defaultBranch":"master","name":"LeetCode-1","ownerLogin":"xiaokai1996","currentUserCanPush":false,"isFork":true,"isEmpty":false,"createdAt":"2019-07-25T07:28:02.000Z","ownerAvatar":"https://avatars.githubusercontent.com/u/44293494?v=4","public":true,"private":false,"isOrgOwned":false},"currentUser":null,"refInfo":{"name":"master","listCacheKey":"v0:1633565888.734648","canEdit":false,"refType":"branch","currentOid":"1756256d7e619164076bbf358c8f7ca68cd8bd79"},"tree":{"items":[{"name":"README.md","path":"098. Recover Binary Search Tree/README.md","contentType":"file"},{"name":"main.cpp","path":"098. Recover Binary Search Tree/main.cpp","contentType":"file"},{"name":"solution.h","path":"098. Recover Binary Search 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可以结合一下 [86. Validate Binary Search Tree](../86. Validate Binary Search Tree), 至少要清楚 BST 的概念。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e就本质而言，BST 的核心特质就是\u003cstrong\u003e有序\u003c/strong\u003e。何谓有序？最直接的方式就是将 BST 按照 Inorder 的遍历思想下落，落下来的序列是有序的。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e3\n/ \u003cbr\u003e\n1 5\n/ \\ / \u003cbr\u003e\n0 2 4 6\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e0 1234 5 6 -\u0026gt; ordered\u003c/p\u003e\n\u003chr\u003e\n\u003cp dir=\"auto\"\u003e回到这道题，让我们 recover 这个 BST, 是因为某一次的 swap 出了错，关键是要理解 swap 出了什么错。提到 two elements, 再结合上面温习的基础概念，可以推测出，这两个 elements 待在了不该待的位置。我们 recover 的目的便是将这两个放错地方的 elements 对调过来。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e题目又说了，不能动 BST 的结构，那么所谓的对调，便是针对值的对调。所以大致可以分为两个步骤：\u003c/p\u003e\n\u003col dir=\"auto\"\u003e\n\u003cli\u003efind two error elements.\u003c/li\u003e\n\u003cli\u003eswap two elements' values.\u003c/li\u003e\n\u003c/ol\u003e\n\u003cp dir=\"auto\"\u003e对于第二步，非常 easy, 我们有 \u003ccode\u003estd::swap\u003c/code\u003e. 但是对于第一条，我们该如何查找，首先就应该想到是 DFS.\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e在 [77. Construct Binary Tree from Inorder and Postorder Traversal](../77. Construct Binary Tree from Inorder and Postorder Traversal) 中我们曾总结，对于二叉树来说，DFS 有三种方式：\u003c/p\u003e\n\u003cul dir=\"auto\"\u003e\n\u003cli\u003e先序遍历（先根遍历）\u003c/li\u003e\n\u003cli\u003e中序遍历（中根遍历）\u003c/li\u003e\n\u003cli\u003e后续遍历（后根遍历）\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp dir=\"auto\"\u003e我们这道题的目的是查出哪里违反了 BST 的次序，正如最开始所说，最直接的方式是通过中序遍历的方式查看序列是否有序。那么我们显然应该选择中序遍历的 DFS 方法。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e写出最简单的中序遍历：\u003c/p\u003e\n\u003cdiv class=\"highlight highlight-source-c++ notranslate position-relative overflow-auto\" dir=\"auto\" data-snippet-clipboard-copy-content=\"void InorderTraversal(TreeNode *root)\n{\n if (root == NULL) return;\n InorderTraversal(root-\u0026gt;left);\n cout \u0026lt;\u0026lt; root-\u0026gt;val \u0026lt;\u0026lt; \u0026quot; \u0026quot;;\n InorderTraversal(root-\u0026gt;right);\n}\"\u003e\u003cpre\u003e\u003cspan class=\"pl-k\"\u003evoid\u003c/span\u003e \u003cspan class=\"pl-en\"\u003eInorderTraversal\u003c/span\u003e(TreeNode *root)\n{\n \u003cspan class=\"pl-k\"\u003eif\u003c/span\u003e (root == \u003cspan class=\"pl-c1\"\u003eNULL\u003c/span\u003e) \u003cspan class=\"pl-k\"\u003ereturn\u003c/span\u003e;\n \u003cspan class=\"pl-c1\"\u003eInorderTraversal\u003c/span\u003e(root-\u0026gt;\u003cspan class=\"pl-smi\"\u003eleft\u003c/span\u003e);\n cout \u0026lt;\u0026lt; root-\u0026gt;\u003cspan class=\"pl-smi\"\u003eval\u003c/span\u003e \u0026lt;\u0026lt; \u003cspan class=\"pl-s\"\u003e\u003cspan class=\"pl-pds\"\u003e\"\u003c/span\u003e \u003cspan class=\"pl-pds\"\u003e\"\u003c/span\u003e\u003c/span\u003e;\n \u003cspan class=\"pl-c1\"\u003eInorderTraversal\u003c/span\u003e(root-\u0026gt;\u003cspan class=\"pl-smi\"\u003eright\u003c/span\u003e);\n}\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e然后，考虑我们的需求，要找到两个元素，而找的条件是次序不符，如何叫次序不符？\u003cstrong\u003e先遍历出来的元素 \u0026gt; 后遍历出来的元素\u003c/strong\u003e，即为不符。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e显然，我们需要三个指针来协助：\u003ccode\u003efirst\u003c/code\u003e, \u003ccode\u003esecond\u003c/code\u003e, \u003ccode\u003eprev\u003c/code\u003e, 分别代表第一个元素，第二个元素，上一个元素。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e当出现 \u003ccode\u003eprev-\u0026gt;val \u0026gt; root-\u0026gt;val\u003c/code\u003e 时，表示找到不符的元素。若是第一个，则为 first. 无论是不是，都将当前的 root 赋给 second, 因为可能会出现仅有两个节点的情况（仅仅遍历一次，second 无法获取应得的值）. 这部分逻辑为：\u003c/p\u003e\n\u003cdiv class=\"highlight highlight-source-c++ notranslate position-relative overflow-auto\" dir=\"auto\" data-snippet-clipboard-copy-content=\"if (first == NULL) first = prev;\nsecond = root;\"\u003e\u003cpre\u003e\u003cspan class=\"pl-k\"\u003eif\u003c/span\u003e (first == \u003cspan class=\"pl-c1\"\u003eNULL\u003c/span\u003e) first = prev;\nsecond = root;\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e然后，便没有太多可说的了。总体来说，本题难度不大。\u003c/p\u003e\n\u003c/article\u003e","errorMessage":null,"headerInfo":{"toc":[],"siteNavLoginPath":"/login?return_to=https%3A%2F%2Fgithub.com%2Fxiaokai1996%2FLeetCode-1%2Ftree%2Fmaster%2F098.%2520Recover%2520Binary%2520Search%2520Tree"}},"totalCount":3,"showBranchInfobar":true},"fileTree":{"":{"items":[{"name":"000. 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