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.…

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 right subtree of a node contains only nodes with keys…

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 right subtree of a node contains only nodes with keys…

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. Consider the following binary search tree: 5 / \ 2 6 / \ 1 3 Example 1: Input: [5,2…

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 right subtree of a node contains only nodes with keys…

给定一个二叉树,判断其是否是一个有效的二叉搜索树.一个二叉搜索树有如下定义:    左子树只包含小于当前节点的数.    右子树只包含大于当前节点的数.    所有子树自身必须也是二叉搜索树.示例 1:    2   / \  1   3二叉树[2,1,3], 返回 true.示例 2:    1   / \  2   3二叉树 [1,2,3], 返回 false.详见:https://leetcode.com/problems/validate-binary-search-tree/descr…

给定一个二叉树,判断其是否是一个有效的二叉搜索树. 假设一个二叉搜索树具有如下特征: 节点的左子树只包含小于当前节点的数. 节点的右子树只包含大于当前节点的数. 所有左子树和右子树自身必须也是二叉搜索树. 示例 1: 输入: 2 / \ 1 3 输出: true 示例 2: 输入: 5 / \ 1 4   / \   3 6 输出: false 解释: 输入为: [5,1,4,null,null,3,6].   根节点的值为 5 ,但是其右子节点值为 4 . class Solution { p…

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…

判断二叉搜索树的方法是: 中序遍历形成递增序列 //全局变量记录中序遍历产生的序列,因为要递归,所以要用全局变量 List<Integer> list = new ArrayList<>(); public boolean isValidBST(TreeNode root) { /* 这里要注意二叉搜索树的左子树的每个节点都要小于根节点,不只是左孩子,右子树也同样, 常犯的一个错误就是只是判断了左孩子和右孩子 正确方法是:中序遍历法,中序遍历之后的结果是一个递增序列 判断是否是二叉…

构建二叉搜索树 /* 利用二叉搜索树的特点:根节点是中间的数 每次找到中间数,左右子树递归子数组 */ public TreeNode sortedArrayToBST(int[] nums) { return builder(nums,0,nums.length-1); } public TreeNode builder(int[] nums,int left,int right) { if (left>right) return null; int mid = (left+right)/2;…