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…

［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］…

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…

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…

Verify Preorder Sequence in Binary Search Tree \Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree. You may assume each number in the sequence is unique. Follow up: Could you do it using on…

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…

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…

Given a binary search tree, print the elements in-order iteratively without using recursion. Note:Before you attempt this problem, you might want to try coding a pre-order traversal iterative solution first, because it is easier. On the other hand, c…

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…

144. Binary Tree Preorder Traversal 前序的非递归遍历:用堆来实现 如果把这个代码改成先向堆存储左节点再存储右节点,就变成了每一行从右向左打印 如果用队列替代堆,并且先存储左节点,再存储右节点,就变成了逐行打印 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> result; if(root == NULL) return res…