[题目] Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following [1,2,2,null,3,null,3] is not: 1 / \ 2 2 \ \ 3…
题面 判断给定二叉树是否对称. Note : empty tree is valid. 算法 1. 根节点判空,若空,则返回true;(空树对称) 2. 根节点不空,递归判断左右子树.如果左右孩子都空,说明到了叶子,返回true;不都空而且一空一不空,返回false:都不空,且值不等,返回false,值相等,递归(左的左, 右的右) && (左的右, 右的左) 源码 /** * Definition for a binary tree node. * struct TreeNode { *…
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following [1,2,2,null,3,null,3] is not: / \ \ \ Note:Bonus po…
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 Note:Bonus points if you could solve it b…
题意:如题 思路:递归解决,同判断对称树的原理差不多.先保证当前两个结点是相等的,再递归保证两左结点是相等的,再递归保证右结点是相等的. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Sol…
101. 对称二叉树 给定一个二叉树,检查它是否是镜像对称的. 例如,二叉树 [1,2,2,3,4,4,3] 是对称的. 1 / \ 2 2 / \ / \ 3 4 4 3 但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的: 1 / \ 2 2 \ \ 3 3 说明: 如果你可以运用递归和迭代两种方法解决这个问题,会很加分. PS: 递归的难点在于:找到可以递归的点 为什么很多人觉得递归一看就会,一写就废. 或者说是自己写无法写出来,关键就是你对递归理解的深不深. 对于…
[101-Symmetric Tree(对称树)] [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, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is…
字典树(查找树) 26个分支作用:检测字符串是否在这个字典里面插入.查找 字典树与哈希表的对比:时间复杂度:以字符来看:O(N).O(N) 以字符串来看:O(1).O(1)空间复杂度:字典树远远小于哈希表 前缀相关的题目字典树优于哈希表字典树可以查询abc是否有ab的前缀 字典树常考点:1.字典树实现2.利用字典树前缀特性解题3.矩阵类字符串一个一个字符深度遍历的问题(DFS+trie) dfs树和trie树同时遍历 word searchIIdfs+hash:时间复杂度大,后面遍历到有些字符就…
LeetCode:Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 算法1:dfs递归的求解 class Solution { public: int minDepth(T…
Description: Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or a…
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…
题目如下: We are given the root node of a maximum tree: a tree where every node has a value greater than any other value in its subtree. Just as in the previous problem, the given tree was constructed from an list A (root = Construct(A)) recursively with…
Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another comput…
前言 在常用的UI组件中,树形组件与数据列表组件可以说是构成一个管理平台基本的两大数据核心组件.树形组件用于系统菜单,数据列表用于数据表现,两者配合即可完成一个简单的数据系统.要实现一个支持复选.工具栏定义的无限级树形组件需要对树的递归创建.树的复选递归有很好的把握,特别是树形在复选情况下的表现样式,这也是难点所在. 树组件需求 (1)支持无限级加载树(支持懒加载方式) (2)支持复选操作,和预定义的复选情况 (3)支持简单树形样式和可自定义叶子点的图片的复杂样式 (4)支持定义工具栏(依赖于t…
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 Note:Bonus points if you could solve it b…
[试题描述]定义一个函数,输入判断一个树是否是另一个对的子树 You have two very large binary trees: T1, with millions of nodes, and T2, with hun-dreds of nodes Create an algorithm to decide if T2 is a subtree of T1 Note that the problem here specifies that T1 has millions of nodes…
方法一,记录子树的上界和下界,root的左子树一定小于root的值,root的右子树一定大于root的值,然后递归左子树和右子树 public class Solution { public boolean isValidBST(TreeNode root) { return isValid(root, null, null); } public boolean isValid(TreeNode root, Integer min, Integer max) { if(root == null)…
对称二叉树的含义非常容易理解,左右子树关于根节点对称,具体来讲,对于一颗对称二叉树的每一颗子树,以穿过根节点的直线为对称轴,左边子树的左节点=右边子树的右节点,左边子树的右节点=左边子树的左节点.所以对称二叉树的定义是针对一棵树,而判断的操作是针对节点,这时可以采取由上到下的顺序,从根节点依次向下判断,只需要重复调用函数,不需要回溯.具体代码如下. class TreeNode: def __init__(self, x): self.val = x self.left = None self.…
LeetCode--Maximum Depth of Binary Tree Question Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Answer /** * Definition for a binary tree…
本篇接着<LeetCode刷题总结-树篇(上)>,讲解有关树的类型相关考点的习题,本期共收录17道题,1道简单题,10道中等题,6道困难题. 在LeetCode题库中,考察到的不同种类的树有七种,分别是二叉搜索树.平衡二叉树.满二叉树.完全二叉树.线段树.字典树和树状数组.每一种类型的树,有着不同的特性以及对应的考察重点.考察重点可参考下图,下文按照树的类型分别划分了一个目录章节,并给出了对应的经典习题. 1 二叉树搜索树 基本定义:又称二叉查找树,二叉排序树.若它的左子树不空,则左子树上所有…
UVa 839 Not so Mobile(树的递归输入) 判断一个树状天平是否平衡,每个测试样例每行4个数 wl,dl,wr,dr,当wl*dl=wr*dr时,视为这个天平平衡,当wl或wr等于0是,下一行将是一个子天平,如果子天平平衡,wl为子天平的wl+wr ,否则整个天平不平衡 #include<iostream> using namespace std; bool solve(int &w) { int wl,dl,wr,dr; cin>>wl>>dl…
[LeetCode]106. Construct Binary Tree from Inorder and Postorder Traversal 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/ 题目描述: Given inorder and postorder traversal…
题目链接:Balanced Binary Tree | LeetCode OJ Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more tha…
Serialize and Deserialize Binary Tree Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstruct…
Quadtrees 四分树就是一颗一个结点只有4个儿子或者没有儿子的树 [题目链接]UVa 297 Quadtrees [题目类型]树的递归 &题意: 一个图片,像素是32*32,给你两个先序递归的字符串,要你把他俩合起来输出面积.p代表结点,f代表黑色,e代表白色. &题解: 用字符串来模拟buf32*32的数组,递归模拟出模型,黑色的涂成1,白色的不处理(也就是0),最后输出1的个数就好了 &代码: #include <bits/stdc++.h> using na…
一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ Given inorder and postorder traversal of a tree, construct the b…
树(tree)[题目描述]从前在森林里面有一棵很大的树,树上住着很多小动物.树上有…
题目 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}, return its bottom-up level order traversal as: 分析 与…
本文讲解有关树的习题中子树问题和新概念定义问题,也是有关树习题的最后一篇总结.前两篇请参考: LeetCode刷题总结-树篇(上) LeetCode刷题总结-树篇(中) 本文共收录9道题,7道中等题,2道困难题.本篇总结的知识点请参考下图: 1 新概念定义问题 本部分收录习题如下: 117.填充每个节点的下一个右侧节点指针II, 难度: 中等 297.二叉树的序列化与反序列化,难度:困难 114.二叉树展开为链表,难度: 中等 998.最大二叉树II, 难度:中等 834.树中距离之和,难度:困…
日常在Web项目开发时,经常会碰到树形架构数据的显示,从数据库中获取数据,并且显示成树形.为了方便,我们可以写一个javascript的一个跨浏览器树控件,后续可以重复使用.本节分享一个自己开发的JS tree插件,对有需要的朋友可以直接下载使用. Tree插件 需要实现 (1).自动将数据组织成上下级关系,这样我们在前端手工添加数据或者从数据库中获取到的数据不需要我们来组织上下级关系 (2).支持自定 加载目录树  支持XML.JSON 格式的数据加载 (3).实现树节点的选择(单选.复选[级…