［抄题］: Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target. [暴力解法]: 时间分析: 空间分析: ［奇葩输出条件］: ［奇葩corner case］: ［思维问题］: 以为要用主函数+ DFS来做.错了,“最近”还是直接用二分法左右查找得了 ［一句话思路］: 定义一个res,如果root离target的距离小 就替换…

Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target. Note: Given target value is a floating point. You are guaranteed to have only one unique value in the BST that is closest to the target.…

①中文题目 给定二叉搜索树(BST)的根节点和一个值. 你需要在BST中找到节点值等于给定值的节点. 返回以该节点为根的子树. 如果节点不存在,则返回 NULL. 例如, 给定二叉搜索树: 在上述示例中,如果要找的值是 5,但因为没有节点值为 5,我们应该返回 NULL. ②思路 这个很好想的, 1.判断当前结点是否为空,如果是的,那就返回null. 2.再看当前结点的值是否跟val相等,如果等,就返回root. 3.如果当前结点的值是否<val,如果是小于,就说明输出的val在当前根的左边,就…

4.5 Implement a function to check if a binary tree is a binary search tree. LeetCode上的原题,请参见我之前的博客Validate Binary Search Tree 验证二叉搜索树.…

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 \ 2 Example 2: Input: [3,1,4,null,null,2] 3 / \ 1 4 / 2…

实现一个二叉搜索树迭代器.你将使用二叉搜索树的根节点初始化迭代器.调用 next() 将返回二叉搜索树中的下一个最小的数.注意: next() 和hasNext() 操作的时间复杂度是O(1),并使用 O(h) 内存,其中 h 是树的高度. 详见:https://leetcode.com/problems/binary-search-tree-iterator/description/ Java实现: /** * Definition for a binary tree node. * publ…

实现一个二叉搜索树迭代器.你将使用二叉搜索树的根节点初始化迭代器. 调用 next() 将返回二叉搜索树中的下一个最小的数. 注意: next() 和hasNext() 操作的时间复杂度是O(1),并使用 O(h) 内存,其中 h 是树的高度. 二叉树的中序遍历 class BSTIterator { public: stack<TreeNode*> s; TreeNode *cur; BSTIterator(TreeNode *root) { cur = root; while(cur) {…

Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target. Note: Given target value is a floating point. You are guaranteed to have only one unique value in the BST that is closest to the target.…

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 only constant space complexity? 这道题让给了我们一个一维数组,让我们…

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…