Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. 解题思路: 先中序遍历找到mistake,然后替换即可,JAVA实现如下: public void recoverTree(TreeNode root) { List<Integer> list = inorderTraversal(root); int left…

题目 Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? 分析 给定一颗二叉排序树,它的两个节点被交换,要求…

二叉排序树中有两个节点被交换了,要求把树恢复成二叉排序树. 详见:https://leetcode.com/problems/recover-binary-search-tree/submissions/ Java实现: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val =…

[LeetCode]99. Recover Binary Search Tree 解题报告(Python) 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/recover-binary-search-tree/description/ 题目描述: Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without chan…

题目链接:Recover Binary Search Tree | LeetCode OJ Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constan…

Recover Binary Search Tree leetcode java https://leetcode.com/problems/recover-binary-search-tree/discuss/32535/No-Fancy-Algorithm-just-Simple-and-Powerful-In-Order-Traversal 描述 解析 解决方法是利用中序遍历找顺序不对的两个点,最后swap一下就好. 因为这中间的错误是两个点进行了交换,所以就是大的跑前面来了,小的跑后面去…

既上篇关于二叉搜索树的文章后,这篇文章介绍一种针对二叉树的新的中序遍历方式,它的特点是不需要递归或者使用栈,而是纯粹使用循环的方式,完成中序遍历. 线索二叉树介绍 首先我们引入“线索二叉树”的概念: "A binary tree is threaded by making all right child pointers that would normally be null point to the inorder successor of the node, and all left chi…

Recover Binary Search Tree Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note:A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? 中序…

Recover Binary Search Tree Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. 互换二叉搜索树中两个位置错误的节点. 思路: 预设三个指针pre,p1,p2.p1用来指向第一个出错的节点,p2用来指向第二个出错的节点. 出错情况有两种,即p1和p2相邻,p1和p2不相邻. 中序遍历此二叉树,用…

Recover Binary Search Tree Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? c…