Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST. Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than or equal to the nod…

题目,本题未做出,还有很多要学习 class Solution { public: vector<int>ans; int base,count,maxCount; void update(int x){ if(base == x){ count++; }else{ base = x; count = 1; } if(count == maxCount) ans.push_back(base); if(count > maxCount) {maxCount = count;ans = v…

501. 二叉搜索树中的众数 给定一个有相同值的二叉搜索树(BST),找出 BST 中的所有众数(出现频率最高的元素). 假定 BST 有如下定义: 结点左子树中所含结点的值小于等于当前结点的值 结点右子树中所含结点的值大于等于当前结点的值 左子树和右子树都是二叉搜索树 例如: 给定 BST [1,null,2,2], 1 \ 2 / 2 返回[2]. 提示:如果众数超过1个,不需考虑输出顺序 进阶:你可以不使用额外的空间吗?(假设由递归产生的隐式调用栈的开销不被计算在内) PS: 遍历 /**…

给定一个有相同值的二叉搜索树(BST),找出 BST 中的所有众数(出现频率最高的元素). 假定 BST 有如下定义: 结点左子树中所含结点的值小于等于当前结点的值 结点右子树中所含结点的值大于等于当前结点的值 左子树和右子树都是二叉搜索树 例如: 给定 BST [1,null,2,2], 1 \ 2 / 2 返回[2]. 提示:如果众数超过1个,不需考虑输出顺序 class Solution { public: map<int, int> check; int MAX = 0; vector…

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST. Basically, the deletion can be divided into two stages: Search for a node to remove. If the n…

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST. Basically, the deletion can be divided into two stages: Search for a node to remove. If the n…

Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST. No…

Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST. No…

230. 二叉搜索树中第K小的元素 题意 给定一个二叉搜索树,编写一个函数 kthSmallest 来查找其中第 k 个最小的元素. 你可以假设 k 总是有效的,1 ≤ k ≤ 二叉搜索树元素个数. 解题思路 中序遍历,利用Python3中提供的生成器方法: 中序遍历,判断存储结点值的数组是否到到k,则表明访问的一个结点就是第k个最小的元素: 先获取跟结点处于的位置(第几个最小的元素),如果它比k小,则从右子结点中找,如果它比k大,则从左子节点中找: 实现 class Solution:    …

      二叉搜索树中第K小的元素     给定一个二叉搜索树,编写一个函数 kthSmallest 来查找其中第 k 个最小的元素. 说明:你可以假设 k 总是有效的,1 ≤ k ≤ 二叉搜索树元素个数. 示例 1: 输入: root = [3,1,4,null,2], k = 1 3 / \ 1 4 \   2 输出: 1 示例 2: 输入: root = [5,3,6,2,4,null,null,1], k = 3 5 / \ 3 6 / \ 2 4 / 1 输出: 3 进阶:如果二叉搜…