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? 使用O(n)空间的话可以直接中序遍历来找问题节点. 如果是…

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…

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…

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? 给定一个排序二叉树,然后任意交换了其中两个节点,要求在使用…

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…

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Example 1: Input: [1,3,null,null,2]   1   /  3   \   2 Output: [3,1,null,null,2]   3   /  1   \   2 Example 2: Input: [3,1,4,null,null…

https://leetcode.com/problems/binary-search-tree-iterator/ Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: nex…

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Example 1: Input: [1,3,null,null,2] 1 / 3 \ 2 Output: [3,1,null,null,2] 3 / 1 \ 2 Example 2: Input: [3,1,4,null,null,2] 3 / \ 1 4 / 2…

原题链接在这里:https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/ 题目: Return the root node of a binary search tree that matches the given preorder traversal. (Recall that a binary search tree is a binary tree where for every…

Leetcode99 给定一个 二叉搜索树,其中两个节点被交换,写一个程序恢复这颗BST. 只想到了时间复杂度O(n)空间复杂度O(h) h为树高的解法,还没想到空间O(1)的解法. 交换的情况只有两种, case1 两个节点在树中(中序遍历序列)相邻, case2 两个节点在树中(中序遍历序列)不相邻. 只要三个指针 Pre first second 即可记录两种case class Solution { public: TreeNode *first; TreeNode *second; T…