题目链接

  题目要求:

  Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

  An example is the root-to-leaf path 1->2->3 which represents the number 123.

  Find the total sum of all root-to-leaf numbers.

  For example,

    1
/ \
2 3 

  The root-to-leaf path 1->2 represents the number 12.
  The root-to-leaf path 1->3 represents the number 13.

  Return the sum = 12 + 13 = 25.

  这道题利用深度优先搜索即可,具体程序(4ms)如下:

 /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode* root) {
int totalSum = ;
sumNumbersSub(root, , totalSum);
return totalSum;
} void sumNumbersSub(TreeNode *tree, int sum, int &totalSum)
{
if(!tree)
return; sum = sum * 10 + tree->val;
if(!tree->left && !tree->right)
{
totalSum += sum;
return;
} sumNumbersSub(tree->left, sum, totalSum);
sumNumbersSub(tree->right, sum, totalSum);
} };

LeetCode之“树”:Sum Root to Leaf Numbers的更多相关文章

  1. 【LeetCode】129. Sum Root to Leaf Numbers 解题报告(Python)

    [LeetCode]129. Sum Root to Leaf Numbers 解题报告(Python) 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/pr ...

  2. LeetCode解题报告—— Sum Root to Leaf Numbers & Surrounded Regions & Single Number II

    1. Sum Root to Leaf Numbers Given a binary tree containing digits from 0-9 only, each root-to-leaf p ...

  3. 【LeetCode】129. Sum Root to Leaf Numbers (2 solutions)

    Sum Root to Leaf Numbers Given a binary tree containing digits from 0-9 only, each root-to-leaf path ...

  4. LeetCode OJ 129. Sum Root to Leaf Numbers

    题目 Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a num ...

  5. 【LeetCode】129. Sum Root to Leaf Numbers

    Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number ...

  6. LeetCode OJ:Sum Root to Leaf Numbers(根到叶节点数字之和)

    Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number ...

  7. 【LeetCode OJ】Sum Root to Leaf Numbers

    # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self ...

  8. LeetCode题解之Sum Root to Leaf Numbers

    1.题目描述 2.问题分析 记录所有路径上的值,然后转换为int求和. 3.代码 vector<string> s; int sumNumbers(TreeNode* root) { tr ...

  9. Leetcode之深度优先搜索(DFS)专题-129. 求根到叶子节点数字之和(Sum Root to Leaf Numbers)

    Leetcode之深度优先搜索(DFS)专题-129. 求根到叶子节点数字之和(Sum Root to Leaf Numbers) 深度优先搜索的解题详细介绍,点击 给定一个二叉树,它的每个结点都存放 ...

  10. LeetCode: Sum Root to Leaf Numbers 解题报告

    Sum Root to Leaf Numbers Given a binary tree containing digits from 0-9 only, each root-to-leaf path ...

随机推荐

  1. Objective-C方法的实现

    Objective-C的方法被两种数据类型描述:一个是选择子(SEL),它用来描述方法的名称;另一个是实现(IMP),它用来描述方法被调用时实际执行的代码(它们基本上只是C函数的指针). 类似于SEL ...

  2. Swift类中如何创建一个对外只读对内可读写的属性

    很简单用private修饰符,后面跟限制关键字set: class Day{ private(set) var rawValue:Int = 0 func showRawValue(){ print( ...

  3. python命令行参数解析模块argparse和docopt

    http://blog.csdn.net/pipisorry/article/details/53046471 还有其他两个模块实现这一功能,getopt(等同于C语言中的getopt())和弃用的o ...

  4. 分析RunTime执行命令以及得到返回值

    RunTime执行命令得到返回值 我们有在好好几篇博客里提到过RunTime,比如 JAVA之旅(二十三)--System,RunTime,Date,Calendar,Math的数学运算 Androi ...

  5. RxJava操作符(05-结合操作)

    转载请标明出处: http://blog.csdn.net/xmxkf/article/details/51656736 本文出自:[openXu的博客] 目录: CombineLatest Join ...

  6. Java使用agent实现main方法之前

    创建Agent项目 PreMainExecutor 类,在main方法之前执行此方法 public class PreMainExecutor { public static void premain ...

  7. JAVA进阶之旅(二)——认识Class类,反射的概念,Constructor,Field,Method,反射Main方法,数组的反射和实践

    JAVA进阶之旅(二)--认识Class类,反射的概念,Constructor,Field,Method,反射Main方法,数组的反射和实践 我们继续聊JAVA,这次比较有意思,那就是反射了 一.认识 ...

  8. Java基本语法-----java数据类型的转换

    前言 Java中可以进行不同数据类型的加减乘除运算吗?是可以的.在算术运算符中已经体验过如果两个整数(int)相除会去掉小数部分.如果需要保留小数部分,可以让除数或者被除数变为double类型的(5变 ...

  9. FORM打开网页链接

     DECLARE l_server_url VARCHAR2(100); l_parameters VARCHAR2(200); BEGIN fnd_profile.get('APPS_WEB_A ...

  10. jvm库对nio的处理

    JVM的IO选择 查JVM源码时刚好看到JVM库的一段代码: public static SelectorProvider create() { String osname = AccessContr ...