既上篇关于二叉搜索树的文章后,这篇文章介绍一种针对二叉树的新的中序遍历方式,它的特点是不需要递归或者使用栈,而是纯粹使用循环的方式,完成中序遍历. 线索二叉树介绍 首先我们引入“线索二叉树”的概念: "A binary tree is threaded by making all right child pointers that would normally be null point to the inorder successor of the node, and all left chi…

二叉搜索树是常用的概念,它的定义如下: 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 greater than the node's key. Both the left and right subtrees must also be binary search…

题目 Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? 分析 给定一颗二叉排序树,它的两个节点被交换,要求…

这是悦乐书的第284次更新,第301篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第152题(顺位题号是669).给定二叉搜索树以及L和R的最低和最高边界,修剪树以使其所有元素位于[L,R](R> = L).可能需要更改树的根,因此结果应返回修剪后的二叉搜索树的新根.例如: 输入:L = 1 R = 2 1 / \ 0 2 输出: 1 \ 2 输入:L = 1 R = 3 3 / \ 0 4 \ 2 / 1 输出: 3 / 2 / 1 本次解题使用的开发工具是ecl…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 解题:果然不能晚上做题,效率好低.看了讨论才学会的解法.设置一个指针next指向当前访问的…

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 解题:应该是很简单的一道题,纠结了好久T_T 基本思路很简单,用栈模拟就可以了.首先根节…

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example:Given binary tree {3,9,20,#,#,15,7}, / \ / \ return its level order traversal as: [ [], [,], [,] ] 题解:二叉树的层次遍历,用队列实现.重点在…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order tr…

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us…

题目链接:Recover Binary Search Tree | LeetCode OJ Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constan…