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. 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…

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…

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. 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…

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…

原题地址 中序遍历二叉搜索树,正常情况下所有元素都应该按递增排列,如果有元素被交换,则会出现前面元素大于后面的情况,称作反序.由于交换了两个节点,所以通常会有两处反序,但如果是两个相邻节点发生了交换,则只会有一次反序. 注意,如果有两次反序,前面那次一定是较大数,后面那次一定是较小数 交换的时候注意只需要交换value就行了,别傻不啦叽的去交换节点. 代码: vector<pair<TreeNode *, TreeNode *> > nodes; TreeNode *prev; v…

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…

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…