437. Path Sum III

Easy

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
package leetcode.easy;

/**
* Definition for a binary tree node. public class TreeNode { int val; TreeNode
* left; TreeNode right; TreeNode(int x) { val = x; } }
*/
public class PathSumIII {
int count = 0; private void helper(TreeNode root, int sum) {
if (root == null) {
return;
} if (root.val == sum) {
count++;
} if (root.left != null) {
helper(root.left, sum - root.val);
} if (root.right != null) {
helper(root.right, sum - root.val);
}
} public int pathSum(TreeNode root, int sum) {
if (root == null) {
return count;
}
helper(root, sum);
if (root.left != null) {
pathSum(root.left, sum);
}
if (root.right != null) {
pathSum(root.right, sum);
}
return count;
} @org.junit.Test
public void test() {
int sum = 8;
TreeNode tn11 = new TreeNode(10);
TreeNode tn21 = new TreeNode(5);
TreeNode tn22 = new TreeNode(-3);
TreeNode tn31 = new TreeNode(3);
TreeNode tn32 = new TreeNode(2);
TreeNode tn34 = new TreeNode(11);
TreeNode tn41 = new TreeNode(3);
TreeNode tn42 = new TreeNode(-2);
TreeNode tn44 = new TreeNode(1);
tn11.left = tn21;
tn11.right = tn22; tn21.left = tn31;
tn21.right = tn32;
tn22.left = null;
tn22.right = tn34; tn31.left = tn41;
tn31.right = tn42;
tn32.left = null;
tn32.right = tn44;
tn34.left = null;
tn34.right = null; tn41.left = null;
tn41.right = null;
tn42.left = null;
tn42.right = null;
tn44.left = null;
tn44.right = null;
System.out.println(pathSum(tn11, sum));
}
}

LeetCode_437. Path Sum III的更多相关文章

  1. 【leetcode】437. Path Sum III

    problem 437. Path Sum III 参考 1. Leetcode_437. Path Sum III; 完

  2. 47. leetcode 437. Path Sum III

    437. Path Sum III You are given a binary tree in which each node contains an integer value. Find the ...

  3. leetcode 112. Path Sum 、 113. Path Sum II 、437. Path Sum III

    112. Path Sum 自己的一个错误写法: class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if(root ...

  4. 437. Path Sum III

    原题: 437. Path Sum III 解题: 思路1就是:以根节点开始遍历找到适合路径,以根节点的左孩子节点开始遍历,然后以根节点的右孩子节点开始遍历,不断循环,也就是以每个节点为起始遍历点 代 ...

  5. 第34-3题:LeetCode437. Path Sum III

    题目 二叉树不超过1000个节点,且节点数值范围是 [-1000000,1000000] 的整数. 示例: root = [10,5,-3,3,2,null,11,3,-2,null,1], sum ...

  6. [LeetCode] Path Sum III 二叉树的路径和之三

    You are given a binary tree in which each node contains an integer value. Find the number of paths t ...

  7. LeetCode 437. Path Sum III (路径之和之三)

    You are given a binary tree in which each node contains an integer value. Find the number of paths t ...

  8. LeetCode算法题-Path Sum III(Java实现)

    这是悦乐书的第227次更新 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第94题(顺位题号是437).您将获得一个二叉树,其中每个节点都包含一个整数值.找到与给定值相加的路径数 ...

  9. [LeetCode] 437. Path Sum III 路径和 III

    You are given a binary tree in which each node contains an integer value. Find the number of paths t ...

随机推荐

  1. 《老子》是帝王术,提倡复古,崇拜圣人,主张愚民,甘居下流,不争上游:4星|李零《人往低处走:<老子>天下第一》

    ​​“ 俗话说,“人往高处走,水往低处流”.<老子>正好相反,它强调的是作“天下谷”.“天下溪”.“天下之牝”,甘居下流,不争上游(第28和第61章).司马谈说,道家的特点是“去健羡,绌聪 ...

  2. Codeforces I. Barcelonian Distance(暴力)

    题目描述: In this problem we consider a very simplified model of Barcelona city. Barcelona can be repres ...

  3. Alpha版本1之后的成绩汇总

    作业要求 1.作业内容: 作业具体要求及评分标准的链接 2.评分细则 •给出开头和团队成员列表(10’) •给出发布地址以及安装手册(20’) •给出测试报告(40’) •给出项目情况总结(30’) ...

  4. centos7部署inotify与rsync实现实时数据同步

    实验环境:CentOS Linux release 7.6.1810 node1:192.168.216.130 客户端(向服务端发起数据同步) node2:192.168.216.132 服务端(接 ...

  5. LOJ P10002 喷水装置 题解

    每日一题 day35 打卡 Analysis 先将不符合条件的区间去掉(即半径小于W,不然宽度无法符合),将符合条件的按区间存入节点中.区间的左边界是x-sqrt(r*r-W*W/4.0),要计算x轴 ...

  6. C# 监控网速

    主要有两个类,其一是NetworkAdapter,该类的作用是获取本机网络适配器列表,并且可以通过该类的属性获取当前网速数据:其二是NetworkMonitor,该类是通过.NET的Performan ...

  7. Linux CentOS7 字符集

    CentOS 7字符集的问题与6有点区别,会出现下面问题,查看是中文,vi进入就变成乱码了 生产中修改配置文件  [root@ce1d2002a999 ~]# cat /etc/locale.conf ...

  8. WinDbg常用命令系列---!analyze

    !analyze命令简介 这个!analyze扩展显示有关当前异常或错误检查的信息. 用户模式: !analyze [-v] [-f | -hang] [-D BucketID] !analyze - ...

  9. Web前端鼠标悬停实现显示与隐藏效果

    css定义,偏移量,相对定位,绝对定位 显示与隐藏 二维码相对于微信图标定位 鼠标悬停微信图标上显示 鼠标离开微信图标时隐藏 什么是定位,就是定义网页标签在运行时显示的位置 css提供Position ...

  10. PHP strtok() 函数

    我们仅在第一次调用 strtok() 函数时使用了 string 参数.在首次调用后,该函数仅需要 split 参数,这是因为它清楚自己在当前字符串中所在的位置. 如需分割一个新的字符串,请再次调用带 ...