Question Given a binary search tree and a node in it, find the in-order successor of that node in the BST. Note: If the given node has no in-order successor in the tree, return null. Solution -- Iterative Inorder result is an ascending array for BST.…

Given a binary search tree and a node in it, find the in-order successor of that node in the BST. Note: If the given node has no in-order successor in the tree, return null. 这道题让我们求二叉搜索树的某个节点的中序后继节点,那么我们根据BST的性质知道其中序遍历的结果是有序的, 是我最先用的方法是用迭代的中序遍历方法,然后用…

原题链接在这里:https://leetcode.com/problems/inorder-successor-in-bst/ Given a binary search tree and a node in it, find the in-order successor of that node in the BST. Note: If the given node has no in-order successor in the tree, return null. Show Company…

Inorder Successor in BST Given a binary search tree and a node in it, find the in-order successor of that node in the BST. Example Given tree = [2,1] and node = 1: 2 / 1 return node 2. Given tree = [2,1,3] and node = 2: 2 / \1 3 return node 3. Note I…

Given a binary search tree and a node in it, find the in-order successor of that node in the BST. The successor of a node p is the node with the smallest key greater than p.val. You will have direct access to the node but not to the root of the tree.…

Given a binary search tree and a node in it, find the in-order successor of that node in the BST. The successor of a node p is the node with the smallest key greater than p.val. Example 1: Input: root = [2,1,3], p = 1 Output: 2 Explanation: 1's in-or…

Given a binary search tree and a node in it, find the in-order successor of that node in the BST. 本题要求查找p在树中的inorder successor(中序遍历时的下一个节点).根据p和root的关系,有三种情况. 1. p在root的左子树中: 1.1 p在左子树中的successor不为空,那么输出这个successor 1.2 p在左子树中的successor为空,那么p的successo…

题目: Given a binary search tree and a node in it, find the in-order successor of that node in the BST. Note: If the given node has no in-order successor in the tree, return null. 链接: http://leetcode.com/problems/inorder-successor-in-bst/ 题解: 一开始的想法就是用…

Given a binary search tree and a node in it, find the in-order successor of that node in the BST. The successor of a node p is the node with the smallest key greater than p.val. Example 1: Input: root = [2,1,3], p = 1 Output: 2 Explanation: 1's in-or…

原题链接在这里:https://leetcode.com/problems/inorder-successor-in-bst-ii/ 题目: Given a binary search tree and a node in it, find the in-order successor of that node in the BST. The successor of a node p is the node with the smallest key greater than p.val. Y…

Given a binary search tree and a node in it, find the in-order successor of that node in the BST. The successor of a node p is the node with the smallest key greater than p.val. Example 1: Input: root = [2,1,3], p = 1 Output: 2 Explanation: 1's in-or…

Given a binary search tree (See Definition) and a node in it, find the in-order successor of that node in the BST. If the given node has no in-order successor in the tree, returnnull. 分析: 给一个二叉查找树,以及一个节点,求该节点的中序遍历后继,如果没有返回 null. 一棵BST定义为: 节点的左子树中的值要严…

Given a binary search tree and a node in it, find the in-order successor of that node in the BST. Note: If the given node has no in-order successor in the tree, return null. Example 1: Input: root = [2,1,3], p = 1 2 / \ 1 3 Output: 2 Example 2: Input…

Question Given a binary search tree, write a function kthSmallest to find the kth smallest element in it. Note: You may assume k is always valid, 1 ≤ k ≤ BST's total elements. Follow up What if the BST is modified (insert/delete operations) often and…

[本文链接] http://www.cnblogs.com/hellogiser/p/query-min-max-successor-of-bst.html [代码]  C++ Code  12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667686970717273747576777879808182…

For the given tree, in order traverse is: visit left side root visit right side // 6,8,10,11,12,15,16,17,20,25,27 The successor is the one right next to the target: // target 8 --> succossor is 10 So, given the tree and target node, to find its succe…

［抄题］: 给一个二叉查找树以及一个节点,求该节点的中序遍历后继,如果没有返回null ［思维问题］: 不知道分合算法和后序节点有什么关系:直接return表达式就行了,它自己会终止的. ［一句话思路］: 比root大时直接扔右边递归,比root小时 考虑是左边递归还是就是root ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: 要定义left节点,留着做后续的比较 ［二刷］: ［三刷］: ［四刷］: ［五刷］: ［总结］…

题意 略 分析 1.首先要了解到BST的中序遍历是递增序列 2.我们用一个临时节点tmp储存p的中序遍历的下一个节点,如果p->right不存在,那么tmp就是从root到p的路径中大于p->val的最小数,否则就遍历p的右子树,找到最左边的节点即可 代码 class Solution { public: /* * @param root: The root of the BST. * @param p: You need find the successor node of p. * @re…

struct node { int val; node *left; node *right; node *parent; node() : val(), left(NULL), right(NULL) { } node(int v) : val(v), left(NULL), right(NULL) { } }; node* insert(int n, node *root) { if (root == NULL) { root = new node(n); return root; } if…

http://www.geeksforgeeks.org/populate-inorder-successor-for-all-nodes/ #include <iostream> #include <vector> #include <algorithm> #include <queue> #include <stack> #include <string> #include <fstream> using namesp…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > re…

Total Accepted: 98729 Total Submissions: 261539 Difficulty: Medium Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could y…

package Tree; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.Deque; import java.util.LinkedList; import java.util.List; import java.util.Queue; import java.util.Stack; class ListNode { int val; Lis…

[94]Binary Tree Inorder Traversal [95]Unique Binary Search Trees II (2018年11月14日,算法群) 给了一个 n,返回结点是 1 - n 的所有形态的BST. 题解:枚举每个根节点 r, 然后递归的生成左右子树的所有集合,然后做笛卡尔积. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * Tr…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…

LinkedIn(39) 1 Two Sum 23.0% Easy 21 Merge Two Sorted Lists 35.4% Easy 23 Merge k Sorted Lists 23.3% Hard 33 Search in Rotated Sorted Array 30.2% Hard 34 Search for a Range 29.1% Medium 46 Permutations 35.7% Medium 47 Permutations II 28.0% Medium 50…

Inorder Successor in BST 要点:这题要注意的是如果不是BST,没法从树结构上从root向那边找p,只能遍历.而根据BST,可以只走正确方向 如果不检查right子树,可以从root到下,但invariant是root!=null.而检查右子树,invariant可以是root!=p 错误点: 不是找到某个>p.val,而是要找到最接近的p.val:所以loop终止条件是直到p==root or root is None,过程中只要>p.val就记录successor:这…

原题链接在这里: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. N…

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…