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 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
- 【leetcode】437. Path Sum III
problem 437. Path Sum III 参考 1. Leetcode_437. Path Sum III; 完
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- [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 ...
- 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 ...
- 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 ...
- [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
- git拉取远程分支并切换到该分支
整理了五种方法,我常用最后一种,这五种方法(除了第4中已经写了fetch的步骤)执行前都需要执行git fetch来同步远程仓库 (1)git checkout -b 本地分支名 origin/远程分 ...
- asp.net core 默认采用小驼峰命名和自定义模型验证
services.AddMvc(options => { options.Filters.Add<ApiExceptionAttribute>(); }).SetCompatibil ...
- Codeforces G. Ant colony
题目描述: F. Ant colonytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputo ...
- poj3522Slim Span(暴力+Kruskal)
思路: 最小生成树是瓶颈生成树,瓶颈生成树满足最大边最小. 数据量较小,所以只需要通过Kruskal,将边按权值从小到大排序,枚举最小边求最小生成树,时间复杂度为O( nm(logm) ) #incl ...
- 浏览器报400-Bad Request异常
今天在使用ie浏览器在测试程序的时候,报这个错误,后台日志打印出来显示的是:连接一个远程主机失败 解决Invalid character found in the request target. Th ...
- LeetCode 1105. Filling Bookcase Shelves
原题链接在这里:https://leetcode.com/problems/filling-bookcase-shelves/ 题目: We have a sequence of books: the ...
- 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 ...
- Xshell6和Xftp6初步使用
Xshell6和Xftp6初步使用 一.Xshell6和Xftp6介绍: Xshell6:可以在Windows界面下用来访问远端不同系统下的服务器,从而比较好的达到远程控制终端的目的. Xftp6:是 ...
- shell脚本编程之变量的小用法
变量赋值 ${parameter:-word}:如果parameter为空或未定义,则变量展开为"word":否则,展开为parameter的值: ${parameter:+wor ...
- Ajax 的一些概念 解析
什么是Ajax Ajax基本概念 Ajax(Asynchronous JavaScript and XML):翻译成中文就是异步的JavaScript和XML. 从功能上来看是一种在无需重新加载整个网 ...