tree traversal
Binary Tree Level Order Traversal
给定一个二叉树,返回其按层次遍历的节点值。 (即逐层地,从左到右访问所有节点)。 例如:给定二叉树: [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 返回其层次遍历结果: [ [3], [9,20], [15,7] ]1、递归版 /** * Definition for a binary
日期 2023-06-12 10:48:40Leetcode 之Binary Tree Postorder Traversal(47)
中序遍历二叉搜索树,得到的是一个有序的结果,找出其中逆序的地方就可以了。如果逆序的地方相邻,只需把逆序的相换即可;如果不相邻,则需要找到第二个逆序对的 第二个元素再做交换。 定义两个指针p和q来指定需要交换的元素,指针pre记录当前结点的前驱结点,用来判断是否逆序。 void recoverTree(TreeNode *root) { pre = p = q
日期 2023-06-12 10:48:40Leetcode 之Binary Tree Postorder Traversal(45)
层序遍历,使用队列将每层压入,定义两个队列来区分不同的层。 vector<vector<int>> levelorderTraversal(TreeNode *root) { vector<vector<int>> result; vector<int>tmp;
日期 2023-06-12 10:48:40Leetcode 之Binary Tree Postorder Traversal(44)
后序遍历,比先序和中序都要复杂。访问一个结点前,需要先判断其右孩子是否被访问过。如果是,则可以访问该结点;否则,需要先处理右子树。 vector<int> postorderTraversal(TreeNode *root) { vector<int> result; stack<TreeNode *>
日期 2023-06-12 10:48:40Leetcode 之Binary Tree Inorder Traversal(43)
树的中序遍历。先不断压入左结点至末尾,再访问,再压入右结点。注意和先序遍历的比较 vector<int> inorderTraversal(TreeNode *root) { vector<int> result; stack<TreeNode *>s; TreeNode *p =
日期 2023-06-12 10:48:40LeetCode:105_Construct Binary Tree from Preorder and Inorder Traversal | 根据前序和中序遍历构建二叉树 | Medium
要求:通过二叉树的前序和中序遍历序列构建一颗二叉树 代码如下: 1 struct TreeNode { 2 int val; 3 TreeNode *left; 4 TreeNode *right; 5 TreeNode(int x): val(x),left(NULL), right(NULL) {} 6 };
日期 2023-06-12 10:48:40LeetCode: 107_Binary Tree Level Order Traversal II | 二叉树自底向上的层次遍历 | Easy
本题和上题一样同属于层次遍历,不同的是本题从底层往上遍历,如下: 代码如下: 1 struct TreeNode { 2 int val; 3 TreeNode* left; 4 TreeNode* right; 5 TreeNode():val(0),left(NULL),right(NULL){} 6
日期 2023-06-12 10:48:40LeetCode: 102_Binary Tree Level Order Traversal | 二叉树自顶向下的层次遍历 | Easy
题目:Binay Tree Level Order Traversal 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,#,
日期 2023-06-12 10:48:40LeetCode:145_Binary Tree Postorder Traversal | 二叉树后序遍历 | Hard
题目:Binary Tree Postorder Traversal 二叉树的后序遍历,题目要求是采用非递归的方式,这个在上数据结构的课时已经很清楚了,二叉树的非递归遍历不管采用何种方式,都需要用到栈结构作为中转,代码很简单,见下: 1 struct TreeNode { 2 int val; 3 TreeNode* left; 4
日期 2023-06-12 10:48:40【LeetCode-面试算法经典-Java实现】【144-Binary Tree Preorder Traversal(二叉树非递归前序遍历)】
【144-Binary Tree Preorder Traversal(二叉树非递归前序遍历)】 【LeetCode-面试算法经典-Java实现】【全部题目文件夹索引】 原题 Given a binary tree, return the preorder traversal of its nodes’ values. For example: G
日期 2023-06-12 10:48:4073_leetcode_Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree 1:中序和后序遍历构成一棵树。2:採用递归的方法。3:把两个数组分别分成两部分;4:注意递归结束情况 TreeNode *buildTree(vector<int> &inorder, vector<
日期 2023-06-12 10:48:40LeetCode之Binary Tree Level Order Traversal 层序遍历二叉树
Binary Tree Level Order Traversal 题目描述: Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example:Given
日期 2023-06-12 10:48:40leetcode第一刷_Construct Binary Tree from Inorder and Postorder Traversal
这道题是为数不多的感觉在读本科的时候见过的问题。人工构造的过程是如何呢。兴许遍历最后一个节点一定是整棵树的根节点。从中序遍历中查找到这个元素,就能够把树分为两颗子树,这个元素左側的递归构造左子树,右側的递归构造右子树。元素本身分配空间,作为根节点。 于set和map容器不同的是。vector容器不含find的成员函数。应该用stl的库函数,好在返回的也是迭代器,而vector的迭代器之间是能
日期 2023-06-12 10:48:401020. Tree Traversals (25)
题目链接:http://www.patest.cn/contests/pat-a-practise/1020 题目: 1020. Tree Traversals (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Suppose that all
日期 2023-06-12 10:48:40Leetcode Binary Tree Zigzag level Order Traversal
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary t
日期 2023-06-12 10:48:40[LeetCode][Java] Binary Tree Level Order Traversal
题目: 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}, 3
日期 2023-06-12 10:48:401086 Tree Traversals Again (25 分)【一般 / 建树 树的遍历】
https://pintia.cn/problem-sets/994805342720868352/problems/994805380754817024 首先要知道的是: 入栈顺序就是树的前序遍历的顺序出栈顺序就是树的
日期 2023-06-12 10:48:40[LeetCode] Binary Tree Zigzag Level Order Traversal
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Gi
日期 2023-06-12 10:48:40[LeetCode] Binary Tree Level Order Traversal II
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,
日期 2023-06-12 10:48:40【LeetCode】145. Binary Tree Postorder Traversal (3 solutions)
Binary Tree Postorder Traversal Given a binary tree, return the postorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3
日期 2023-06-12 10:48:40[LeetCode] 1028. Recover a Tree From Preorder Traversal 从先序遍历还原二叉树
We run a preorder depth first search on the rootof a binary tree. At each node in this traversal, we output D dashes (where D is the depth of this node),
日期 2023-06-12 10:48:40[LeetCode] 94. Binary Tree Inorder Traversal 二叉树的中序遍历
Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution
日期 2023-06-12 10:48:40[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历建立二叉树
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorde
日期 2023-06-12 10:48:4094. Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is triv
日期 2023-06-12 10:48:40leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 从中序与后序遍历序列构造二叉树(中等)
一、题目大意 给定两个整数数组 inorder 和 postorder ,其中 inorder 是二叉树的中序遍历, postorder 是同一棵树的后序遍历,请你构造并返回这颗
日期 2023-06-12 10:48:40leetcode 889. Construct Binary Tree from Preorder and Postorder Traversal 根据前序和后
一、题目大意 给定两个整数数组,preorder 和 postorder ,其中 preorder 是一个具有 无重复 值的二叉树的前序遍历,postorder 是同一棵树的后序遍
日期 2023-06-12 10:48:40