【回溯/二叉树】113. Path Sum II

113. 路径总和 II(sum path ii) 给定一个二叉树和一个目标和,找到所有从根节点到叶子节点路径总和等于给定目标和的路径。 说明:叶子节点是指没有子节点的节点。 示例: 给定如下二叉树,以及目标和sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ ... 阅读详情

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> ans;
        vector<int> temp;
        if(root==NULL)
    		return ans;
        find(ans,temp,root,0,sum);
        return ans;
    }
    void find(vector<vector<int>> &ans,vector<int> temp,TreeNode *root,int sum,int target)
    {
    	temp.push_back(root->val);
    	if(root->left==NULL&&root->right==NULL)
    	{
    		sum += root->val;
    		if(target==sum)
    			ans.push_back(temp);
    		return;
    	}

    	if(root->left)
    	{
    		find(ans,temp,root->left,sum+root->val,target);
    	}
    	if(root->right)
    		find(ans,temp,root->right,sum+root->val,target);

    }
};

 

 

 

嵌入式学习笔记——EEPROM(M24M01) M24M01是ST公司出品的1Mbit(128Kbyte)I2C兼容EEPROM,兼容所有I2C总线模式(1MHz、400kHz、100kHz),Page大小为256字节,支持硬件写保护,超过400万次写入操作,超过200年的数据留存通过模块化设计,涵盖了 M24M01-RMN6P 的功能介绍、引脚架构、硬件连线、模块化代码以及性能优化思路,我们将 M24M01 EEPROM 与 STM32 的集成变得高效而可靠。希望能帮你在 STM32 上快速上手,并让你的项目存储管理更高效可靠。 阅读详情

相关推荐

NIDS——suricata(一)

NIDS——suricata环境安装和规则编写

weixin_58666006的博客 1273

数据结构(六)—— 二叉树(4)回溯

回溯

weixin_42690036的博客 1459

Android14 RK3576开发板有线网卡静态IP配置全攻略(附数据库修改详解)

本文详细介绍了在Android 14 RK3576开发板上实现有线网卡出厂默认静态IP配置的完整方案。通过深入修改系统源码和SettingsProvider数据库,扩展全局设置字段并注入默认值,确保设备首次启动即自动应用预设的IP地址、子网掩码、网关和DNS,适用于工业控制、数字标牌等批量部署场景,有效提升生产效率。

weixin_29253711的博客 609

二叉树--递归和回溯

设计一个递归函数,该函数的作用是计算包含当前节点的最大路径和。

MADAO的博客 246

Nexus3国内高速下载指南:多版本全平台资源汇总

本文针对国内开发者从官网下载Nexus 3速度慢、易失败的问题,提供了详尽的解决方案。汇总了Windows、Linux、macOS等多个平台的Nexus 3历史及最新版本(如3.31.1)的国内高速下载链接,并附有版本选择建议与Linux环境下的快速启动指南,帮助用户高效获取和部署这一流行的仓库管理工具。

jwt8token的博客 435

EV1527、PT2262傻傻分不清?一文讲透常见315/433MHz遥控编码协议区别与选型

本文深度对比了EV1527、PT2262和SC5262三种常见的315/433MHz遥控编码协议,详细解析了它们的技术特点、成本效益和适用场景。EV1527作为低成本方案的首选,特别适合智能家居和简易安防设备;PT2262在稳定性和兼容性上表现优异;而SC5262则在安全性方面具有显著优势。文章还提供了实用的选型框架和工程建议,帮助开发者在不同应用场景中做出最优决策。

weixin_33640562的博客 354

1022. 从根到叶的二进制数之和

package com.heu.wsq.leetcode.tree; /** * 1022. 从根到叶的二进制数之和 * @author wsq * @date 2020/12/22 * 给出一棵二叉树,其上每个结点的值都是 0 或 1 。每一条从根到叶的路径都代表一个从最高有效位开始的二进制数。例如,如果路径为 0 -> 1 -> 1 -> 0 -> 1,那么它表示二进制数 01101,也就是 13 。 * 对树上的每一片叶子,我们都要找出从根到该叶子的路径所表示的数字

Codeoh的博客 281

从零开始:用Neo4j构建你的第一个知识图谱(附完整代码示例)

本文提供了一份从零开始的Neo4j知识图谱构建实战指南。通过一个“经典科幻电影宇宙”的具体案例,手把手教你安装环境、设计图谱模式,并使用Cypher查询语言完成节点与关系的创建、数据查询及可视化,帮助你快速掌握自下而上构建知识图谱的核心技能。

weixin_29053073的博客 475

二叉树搜索序列(回溯)

回溯算法

少渊的博客 1292

VHDL六层电梯控制器及仿真

六层电梯电梯 设计要求:设计一个六层电梯控制系统要求有超载、关门中断、提前关门清除报警信电梯外人的上升请求信号,电梯外人的下降请求信号, 电梯外人上升请求信号显示电梯外人下降请求信号显示到达楼层信号 设计思路:输入有clk时钟信号(频率为2Hz) full,deng,quick,clr超载、关门中断、提前关门清除报警c_u1,c_u2,c_u3,c_u4,c_u5电梯外人的上升请求c_d2,c_d3,c_d4,c_d5,c_d6 电梯外人的下降请求 d1,d2,d3,d4,d5,d6电梯内人的...

QQ_778132974的博客 1712

Python爬虫教程:抓取区块链交易信息及加密货币市场数据

随着区块链技术和加密货币的迅猛发展,区块链交易和加密货币市场的数据逐渐成为金融、技术、经济研究等领域的热点。对于开发者和研究者而言,实时获取区块链交易数据和加密货币市场行情,对于投资分析、市场预测、技术研究等具有重要的参考价值。本文将通过Python爬虫技术,介绍如何抓取区块链交易信息及加密货币市场数据,详细阐述数据获取的原理、技术方案、实现方法以及抓取到的数据的存储与分析。我们将依托最新的爬虫技术,如requestsSeleniumScrapy,以及相关的数据处理库(如pandasmatplotlib。

2201_76125261的博客 2778

【二叉树】【回溯】二叉树的所有路径详解【力扣.257】超详细的宝藏教程

【二叉树】【回溯】二叉树的所有路径详解【力扣.257】超详细的宝藏教程 干货满满,不收藏绝对后悔!

系统编程、网络编程 2783

[Dify] 实现“多知识库切换”功能的最佳实践

知识库切换是提升AI问答系统精准度的关键技术。本文介绍了Dify平台实现多知识库切换的三种方式:静态配置、元数据过滤和动态切换,分析了各自的优缺点。最佳实践建议包括:合理分库设计、意图分类、主库优先检索等。文章还提供了示例工作流程和注意事项,帮助构建更智能、精准的知识驱动型问答系统。通过合理切换知识库,可显著提升回答相关性和系统性能。

技术小甜甜:传播可落地技术,帮助个人和企业降低解决真实场景问题的成本 442

LeetCode 113. Path Sum II 题解

本题是二叉树的经典问题,主要考察对二叉树遍历和回溯的理解和应用。通过使用深度优先搜索和回溯,我们可以高效地找到所有从根节点到叶子节点路径总和等于给定目标和的路径。深度优先搜索的核心思想是:遍历二叉树,记录当前路径和当前路径的和,当到达叶子节点时,检查路径和是否等于目标和。回溯的核心思想是:在递归返回时移除当前节点,以便探索其他路径。这种方法不仅适用于路径总和 II 问题,还可以应用于许多其他需要遍历二叉树并记录路径的场景。掌握深度优先搜索和回溯的思想,对于解决这类问题非常重要。

cannonjinx的博客 2325

TMS_WebGMaps_1.9.8.0_Full_Source

TMS_WebGMaps_1.9.8.0_Full_Source 全源码 delphi cbuilder

jre9下载 jdk9

(Java SE Runtime Environment 9.0) 9.0u175 官方最新版 【x86|x64】

leetcode 113. Path Sum II-路径和|回溯算法

原题链接:113. Path Sum II 【思路】 采用回溯算法。回溯算法要求我们准备一个中间集 temp,一个结果集 res。这里和以往题目递归中包含循环,循环中嵌套递归的不同。本题没有包含循环。 递归遍历过程中,往中间集当中添加节点值,遍历结束的条件为 root 的左右节点均为空,即到叶子节点为止,如果满足条件——路径上各节点值和等于 sum,那么就可以将结果集添加到中间集当中。回溯是在

hdtopku 2557

深度学习工具包 DeepLearningToolbox

深度学习工具包Deprecation notice.-----This toolbox is outdated and no longer maintained.There are much better tools available for deep learning than this toolbox, e.g. [Theano](http://deeplearning.net/software/theano/), [torch](http://torch.ch/) or [tensorflow](http://www.tensorflow.org/)I would suggest you use one of the tools mentioned above rather than use this toolbox.Best, Rasmus.DeepLearnToolbox================A Matlab toolbox for Deep Learning.Deep Learning is a new subfield of machine learning that focuses on learning deep hierarchical models of data.It is inspired by the human brain's apparent deep (layered, hierarchical) architecture.A good overview of the theory of Deep Learning theory is[Learning Deep Architectures for AI](http://www.iro.umontreal.ca/~bengioy/papers/ftml_book.pdf)For a more informal introduction, see the following videos by Geoffrey Hinton and Andrew Ng.* [The Next Generation of Neural Networks](http://www.youtube.com/watch?v=AyzOUbkUf3M) (Hinton, 2007)* [Recent Developments in Deep Learning](http://www.youtube.com/watch?v=VdIURAu1-aU) (Hinton, 2010)* [Unsupervised Feature Learning and Deep Learning](http://www.youtube.com/watch?v=ZmNOAtZIgIk) (Ng, 2011)If you use this toolbox in your research please cite [Prediction as a candidate for learning deep hierarchical models of data](http://www2.imm.dtu.dk/pubdb/views/publication_details.php?id=6284)```@MASTERSTHESIS\{IMM2012-06284, author = "R. B. Palm", title = "Prediction as a candidate for learning deep hierarchical models of data", year = "2012",}```Contact: rasmusbergpalm at gmail dot comDirectories included in the toolbox-----------------------------------`NN/` - A library for Feedforward Backpropagation Neural Networks`CNN/` - A library for Convolutional Neural Networks`DBN/` - A library for Deep Belief Networks`SAE/` - A library for Stacked Auto-Encoders`CAE/` - A library for Convolutional Auto-Encoders`util/` - Utility functions used by the libraries`data/` - Data used by the examples`tests/` - unit tests to verify toolbox is workingFor references on each library check REFS.mdSetup-----1. Download.2. addpath(genpath('DeepLearnToolbox'));Example: Deep Belief Network---------------------```matlabfunction test_example_DBNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);%% ex1 train a 100 hidden unit RBM and visualize its weightsrand('state',0)dbn.sizes = [100];opts.numepochs = 1;opts.batchsize = 100;opts.momentum = 0;opts.alpha = 1;dbn = dbnsetup(dbn, train_x, opts);dbn = dbntrain(dbn, train_x, opts);figure; visualize(dbn.rbm{1}.W'); % Visualize the RBM weights%% ex2 train a 100-100 hidden unit DBN and use its weights to initialize a NNrand('state',0)%train dbndbn.sizes = [100 100];opts.numepochs = 1;opts.batchsize = 100;opts.momentum = 0;opts.alpha = 1;dbn = dbnsetup(dbn, train_x, opts);dbn = dbntrain(dbn, train_x, opts);%unfold dbn to nnnn = dbnunfoldtonn(dbn, 10);nn.activation_function = 'sigm';%train nnopts.numepochs = 1;opts.batchsize = 100;nn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.10, 'Too big error');```Example: Stacked Auto-Encoders---------------------```matlabfunction test_example_SAEload mnist_uint8;train_x = double(train_x)/255;test_x = double(test_x)/255;train_y = double(train_y);test_y = double(test_y);%% ex1 train a 100 hidden unit SDAE and use it to initialize a FFNN% Setup and train a stacked denoising autoencoder (SDAE)rand('state',0)sae = saesetup([784 100]);sae.ae{1}.activation_function = 'sigm';sae.ae{1}.learningRate = 1;sae.ae{1}.inputZeroMaskedFraction = 0.5;opts.numepochs = 1;opts.batchsize = 100;sae = saetrain(sae, train_x, opts);visualize(sae.ae{1}.W{1}(:,2:end)')% Use the SDAE to initialize a FFNNnn = nnsetup([784 100 10]);nn.activation_function = 'sigm';nn.learningRate = 1;nn.W{1} = sae.ae{1}.W{1};% Train the FFNNopts.numepochs = 1;opts.batchsize = 100;nn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.16, 'Too big error');```Example: Convolutional Neural Nets---------------------```matlabfunction test_example_CNNload mnist_uint8;train_x = double(reshape(train_x',28,28,60000))/255;test_x = double(reshape(test_x',28,28,10000))/255;train_y = double(train_y');test_y = double(test_y');%% ex1 Train a 6c-2s-12c-2s Convolutional neural network %will run 1 epoch in about 200 second and get around 11% error. %With 100 epochs you'll get around 1.2% errorrand('state',0)cnn.layers = { struct('type', 'i') %input layer struct('type', 'c', 'outputmaps', 6, 'kernelsize', 5) %convolution layer struct('type', 's', 'scale', 2) %sub sampling layer struct('type', 'c', 'outputmaps', 12, 'kernelsize', 5) %convolution layer struct('type', 's', 'scale', 2) %subsampling layer};cnn = cnnsetup(cnn, train_x, train_y);opts.alpha = 1;opts.batchsize = 50;opts.numepochs = 1;cnn = cnntrain(cnn, train_x, train_y, opts);[er, bad] = cnntest(cnn, test_x, test_y);%plot mean squared errorfigure; plot(cnn.rL);assert(er<0.12, 'Too big error');```Example: Neural Networks---------------------```matlabfunction test_example_NNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);% normalize[train_x, mu, sigma] = zscore(train_x);test_x = normalize(test_x, mu, sigma);%% ex1 vanilla neural netrand('state',0)nn = nnsetup([784 100 10]);opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samples[nn, L] = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.08, 'Too big error');%% ex2 neural net with L2 weight decayrand('state',0)nn = nnsetup([784 100 10]);nn.weightPenaltyL2 = 1e-4; % L2 weight decayopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex3 neural net with dropoutrand('state',0)nn = nnsetup([784 100 10]);nn.dropoutFraction = 0.5; % Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

113. Path Sum II**

113. Path Sum II** https://leetcode.com/problems/path-sum-ii/ 题目描述 Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum. Note: A leaf is a node with no...

珍妮的选择的博客 262

visio安装包

Microsoft Office Visio 是一款专业的制图软件。利用经过更新的形状、协作工具和数据链接图表,创建专业级图表以简化复杂的信息。

LeetCode: 113. Path Sum II

这道题折磨了我好长时间,终于通过了所有test。语言c++,时间16ms。成就感能让我继续努力,所以记录一下。题目在这里Path Sum II,题目的意思是定义从二叉树的根结点到叶子结点为一条路径,将路径上所有结点的值加起来是该路径的和(path sum),现在给定一个整数,求出所有路径和等于该数的路径,结果保存在二维数组中。思路一:递归。与二叉树有关的大部分问题都可以用递归来处理,当问题规模不大时

青春不谢的专栏 975

[LeetCode] Path Sum I/II/III 回溯算法全解

[LeetCode] Path Sum I/II/III 回溯算法全解 回溯算法的框架: 参考文章. 解决一个回溯问题,实际上就是一个决策树的遍历过程。你只需要思考 3 个问题: 1、路径:也就是已经做出的选择。 2、选择列表:也就是你当前可以做的选择。 3、结束条件:也就是到达决策树底层,无法再做选择的条件。 result = [] def backtrack(路径, 选择列表): if 满足结束条件: result.add(路径) return for

xvrixingkong的博客 205

【LeetCode】113. Path Sum II 解题报告(Python)

【LeetCode】113. Path Sum II 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/path-sum-ii/descript...

负雪明烛 2165

[LeetCode]113.Path Sum II

【题目】 Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22, 5

SmartSi 2044

113. Path Sum II(路径和II)

原题链接:https://leetcode.com/problems/path-sum-ii/ Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. Note:A leaf is a node with no children. Examp...

God_Mood的博客 176

leetcode解题之112 & 113 & 437. Path Sum java版(二叉树路径和)

112. Path Sum ,113. Path Sum II,437. Path Sum III, leetcode解题之112 & 113 & 437. Path Sum java版(二叉树路径和)

mine_song的博客 1114

Leetcode113. 路径总和 Path Sum II(Java)

Leetcode113. 路径总和 Path Sum II(Java) ##Tree##, ##Depth-first Search## 路径总和II 参考Leetcode112路径总和,用回溯法 因为需要找到所有路径,因此需要DFS整棵子树 dfs(TreeNode root, int sum)表示找到所有从root出发,到叶节点和为sum的路径的函数 递归遍历左右儿子结点,以左儿子结点为例dfs(root.left, sum - root.val)表示找到所有从root.left出发,到叶子节点和为s

qq_42411921的博客 206

【leetcode】113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. Note: A leaf is a node with no children. Example: Given the below binary tree and sum = 22, ...

ghscarecrow的博客 267

LeetCode | 0113. Path Sum II路径总和 II【Python】

LeetCode 0113. Path Sum II路径总和 II【Medium】【Python】【回溯】

Wonz 586

Path Sum II 二叉树路径和,打印所有路径 @LeetCode

经典递归回溯! package Level2; import java.util.ArrayList; import Utility.TreeNode; /** * Path Sum II * * Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equa

4831
上一篇: 【Array】73. Set Matrix Zeroes
下一篇: see-14. Longest Common Prefix
云水谣CS
博客等级 码龄9年 8粉丝 268原创
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付:10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值