C++实现LeetCode(173.二叉搜索树迭代器)

[LeetCode] 173.Binary Search Tree Iterator 二叉搜索树迭代器

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: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

这道题主要就是考二叉树的中序遍历的非递归形式，需要额外定义一个栈来辅助，二叉搜索树的建树规则就是左<根<右，用中序遍历即可从小到大取出所有节点。代码如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    BSTIterator(TreeNode *root) {
        while (root) {
            s.push(root);
            root = root->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !s.empty();
    }

    /** @return the next smallest number */
    int next() {
        TreeNode *n = s.top();
        s.pop();
        int res = n->val;
        if (n->right) {
            n = n->right;
            while (n) {
                s.push(n);
                n = n->left;
            }
        }
        return res;
    }
private:
    stack<TreeNode*> s;
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */
赞 (0)

相关推荐

  • C++实现LeetCode165.版本比较)

    [LeetCode] 165.Compare Version Numbers 版本比较 Compare two version numbers version1 and version2. If version1 > version2 return 1; if version1 <version2 return -1;otherwise return 0. You may assume that the version strings are non-empty and contain onl

  • C++实现LeetCode(171.求Excel表列序号)

    [LeetCode] 171.Excel Sheet Column Number 求Excel表列序号 Related to question Excel Sheet Column Title Given a column title as appear in an Excel sheet, return its corresponding column number. For example:     A -> 1 B -> 2 C -> 3 ... Z -> 26 AA -&g

  • C++实现LeetCode(172.求阶乘末尾零的个数)

    [LeetCode] 172. Factorial Trailing Zeroes 求阶乘末尾零的个数 Given an integer n, return the number of trailing zeroes in n!. Example 1: Input: 3 Output: 0 Explanation: 3! = 6, no trailing zero. Example 2: Input: 5 Output: 1 Explanation: 5! = 120, one trailing

  • C++实现LeetCode(168.求Excel表列名称)

    [LeetCode] 168.Excel Sheet Column Title 求Excel表列名称 Given a positive integer, return its corresponding column title as appear in an Excel sheet. For example:     1 -> A 2 -> B 3 -> C ... 26 -> Z 27 -> AA 28 -> AB ... Example 1: Input: 1 O

  • C++实现LeetCode(166.分数转循环小数)

    [LeetCode] 166.Fraction to Recurring Decimal 分数转循环小数 Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If the fractional part is repeating, enclose the repeating part in parentheses. Fo

  • C++实现LeetCode(169.求大多数)

    [LeetCode] 169. Majority Element 求大多数 Given an array nums of size n, return the majority element. The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array. Example 1

  • C++实现LeetCode(170.两数之和之三 - 数据结构设计)

    [LeetCode] 170. Two Sum III - Data structure design 两数之和之三 - 数据结构设计 Design and implement a TwoSum class. It should support the following operations: add and find. add - Add the number to an internal data structure. find - Find if there exists any pai

  • C++实现LeetCode(167.两数之和之二 - 输入数组有序)

    [LeetCode] 167.Two Sum II - Input array is sorted 两数之和之二 - 输入数组有序 Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of t

  • C++实现LeetCode(173.二叉搜索树迭代器)

    [LeetCode] 173.Binary Search Tree Iterator 二叉搜索树迭代器 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: next() and

  • C++实现LeetCode(96.独一无二的二叉搜索树)

    [LeetCode] 96. Unique Binary Search Trees 独一无二的二叉搜索树 Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n? Example: Input: 3 Output: 5 Explanation: Given n = 3, there are a total of 5 unique BST's: 1         3  

  • C++实现LeetCode(95.独一无二的二叉搜索树之二)

    [LeetCode] 95. Unique Binary Search Trees II 独一无二的二叉搜索树之二 Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n. Example: Input: 3 Output: [ [1,null,3,2], [3,2,null,1], [3,1,null,null,2], [2,1,3],

  • C++实现LeetCode(99.复原二叉搜索树)

    [LeetCode] 99. Recover Binary Search Tree 复原二叉搜索树 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

  • C++实现LeetCode(98.验证二叉搜索树)

    [LeetCode] 98. Validate Binary Search Tree 验证二叉搜索树 Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The ri

  • C++实现LeetCode(109.将有序链表转为二叉搜索树)

    [LeetCode] 109.Convert Sorted List to Binary Search Tree 将有序链表转为二叉搜索树 Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. For this problem, a height-balanced binary tree is defined as a binary

  • C++实现LeetCode(108.将有序数组转为二叉搜索树)

    [LeetCode] 108.Convert Sorted Array to Binary Search Tree 将有序数组转为二叉搜索树 Given an array where elements are sorted in ascending order, convert it to a height balanced BST. For this problem, a height-balanced binary tree is defined as a binary tree in wh

  • Java数据结构之二叉搜索树详解

    Java数据结构之二叉搜索树详解

    目录 前言 性质 实现 节点结构 初始化 插入节点 查找节点 删除节点 最后 前言 今天leetcode的每日一题450是关于删除二叉搜索树节点的,题目要求删除指定值的节点,并且需要保证二叉搜索树性质不变,做完之后,我觉得这道题将二叉搜索树特性凸显的很好,首先需要查找指定节点,然后删除节点并且保持二叉搜索树性质不变,就想利用这个题目讲讲二叉搜索树. 二叉搜索树作为一个经典的数据结构,具有链表的快速插入与删除的特点,同时查询效率也很优秀,所以应用十分广泛,例如在文件系统和数据库系统一般会采用这种数

  • Java动态规划方式解决不同的二叉搜索树

    Java动态规划方式解决不同的二叉搜索树

    目录 一.题目描述 二.思路 三.代码 一.题目描述 给你一个整数 n ,求恰由 n 个节点组成且节点值从 1 到 n 互不相同的 二叉搜索树 有多少种?返回满足题意的二叉搜索树的种数. 来源:https://leetcode.cn/problems/unique-binary-search-trees/ 二.思路 本题可以使用动态规划的方式解决,我们先来看一下大题思路.以 n = 3 为例,n = 3 时的不同的二叉搜索树数目,可以通过分别 以 1 为根节点,以 2 为根节点,以 3 为根节点

  • Javascript实现从小到大的数组转换成二叉搜索树

    废话不多说了,直接给大家贴代码了,具体代码如下所示: var Array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; var Tree = createTree(Array); console.log(Tree); // 构造一个节点 function Node(nodeData, leftData, rightData) { this.nodeData = nodeData; this.leftData = leftData; this.rightData = rig

随机推荐