Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL. For example, Given the tree: 4 / \ ...
// 借助中序遍历,左根右的顺序publicTreeNode searchBST(TreeNode root, intval) {// 先判空if(root ==null) {returnnull; }// 如果val大于当前节点值,就往右子树里面找if(root.val<val) {returnsearchBST(root.right,val); }// 如果相等,返回当前以节点为根节点的子树if(root.val==val) {returnroot...
给定二叉搜索树(BST)的根节点和要插入树中的值,将值插入二叉搜索树。返回插入后二叉搜索树的根节点。保证原始二叉搜索树中不存在新值。 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...
Left } } /** * Your BSTIterator object will be instantiated and called as such: * obj := Constructor(root); * param_1 := obj.Next(); * param_2 := obj.HasNext(); */ 题目链接: Binary Search Tree Iterator : leetcode.com/problems/b 二叉搜索树迭代器: leetcode-cn.com/problem ...
【LeetCode】11. Lowest Common Ancestor of a Binary Search Tree· 二叉搜索树的最近公共祖先 秦她的菜 吉利 程序员 1 人赞同了该文章 题目描述 英文版描述 Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.According to the definition of...
leetcode 669. Trim a Binary Search Tree 修建二叉搜索树BST + 深度优先遍历DFS,Givenabinarysearchtreeandthelowestandhighestboundariesas
LeetCode 每日一题 Daily Challenge 107 Binary Tree Level Order Traversal II 369 -- 4:05 App LeetCode 每日一题 Daily Challenge 239 Sliding Window Maximum 113 1 2:43 App LeetCode 每日一题 Daily Challenge 538 Convert BST to Greater Tree 237 -- 5:28 App LeetCode 每日一题 Daily Challenge 79...
LeetCode-701. Insert into a Binary Search Tree 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 ...
Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. 调用next()将返回二叉搜索树中的下一个最小的数。 Callingnext()will return the next smallest number in the BST.
今天介绍的是LeetCode算法题中Easy级别的第163题(顺位题号是700)。给定一个二叉搜索树(BST)的和正整数val。 你需要在BST中找到节点的值等于给定val的节点。返回以该节点为根的子树。如果此节点不存在,则应返回null。例如: 鉴于树: 4 / \ 2 7 / \ 1 3 ...