binary-tree-maximum-path-sum(mock)
注意:
// 注意,如果一个类放在另一个类里面,初始化时候会报错 Solution is not a enclosing class
// 这是因为如果TreeNode不是static,那么要求先有外部类的实例
// 要加上static
// 或者放到类的外面
https://leetcode.com/problems/binary-tree-maximum-path-sum/
https://leetcode.com/mockinterview/session/result/xslp8c2/
package com.company; import java.util.ArrayList;
import java.util.List; class Solution { // 注意,如果放在Solution里面,会报错 Solution is not a enclosing class
// 这是因为如果TreeNode不是static,那么要求先有外部类的实例
// 要加上static
static class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
} public int maxPathSum(TreeNode root) {
// 要求至少有一个元素,全是负数情况下不能认为是0
if (root == null) {
return 0;
}
List<Integer> ret = impl(root);
return ret.get(1);
} // 多值返回一般放在容器里
// [0]包含root的单一路径最大值; [1] 最大值;
// 调用时保证root不会为null
private List<Integer> impl(TreeNode root) {
List<Integer> ret = new ArrayList();
int maxWithRoot = root.val;
int maxRet = root.val; if (root.left != null) {
List<Integer> left = impl(root.left);
System.out.printf("Here is left %d, ret: %d, %d\n", root.left.val, left.get(0), left.get(1));
if (left.get(0) > 0) {
maxWithRoot = root.val + left.get(0);
}
maxRet = maxWithRoot > left.get(1) ? maxWithRoot : left.get(1); }
if (root.right != null) {
List<Integer> right = impl(root.right);
int tmp = maxWithRoot;
if (root.val + right.get(0) > maxWithRoot) {
maxWithRoot = root.val + right.get(0);
}
// 下面这个地方因为考虑不周,导致了一个bug,只考虑了maxWithRoot,没有考虑之前的maxRet
maxRet = maxWithRoot > maxRet ? maxWithRoot : maxRet;
maxRet = maxRet > right.get(1) ? maxRet : right.get(1); // merge two branch
if (tmp + right.get(0) > maxRet) {
maxRet = tmp + right.get(0);
}
} ret.add(maxWithRoot);
ret.add(maxRet);
System.out.printf("Here is node %d, ret: %d, %d\n", root.val, maxWithRoot, maxRet);
return ret;
}
} public class Main { public static void main(String[] args) {
// write your code here
System.out.println("Hello"); Solution.TreeNode node1 = new Solution.TreeNode(1);
Solution.TreeNode node2 = new Solution.TreeNode(2);
Solution.TreeNode node3 = new Solution.TreeNode(3);
Solution.TreeNode node4 = new Solution.TreeNode(4);
Solution.TreeNode node5 = new Solution.TreeNode(5);
Solution.TreeNode node6 = new Solution.TreeNode(6);
Solution.TreeNode node7 = new Solution.TreeNode(7);
Solution.TreeNode node8 = new Solution.TreeNode(8);
Solution.TreeNode node9 = new Solution.TreeNode(9);
Solution.TreeNode node10 = new Solution.TreeNode(10);
node1.left = node2;
node1.right = node3;
node2.left = node4;
node2.right = node5;
node3.left = node6;
node3.right = node7;
node4.left = node8;
node4.right = node9;
node5.left = node10; Solution solution = new Solution();
int ret = solution.maxPathSum(node1);
System.out.printf("Get ret: %d\n", ret); }
}
binary-tree-maximum-path-sum(mock)的更多相关文章
- [leetcode]Binary Tree Maximum Path Sum
Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and ...
- 【leetcode】Binary Tree Maximum Path Sum
Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and ...
- 26. Binary Tree Maximum Path Sum
Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and ...
- leetcode 124. Binary Tree Maximum Path Sum 、543. Diameter of Binary Tree(直径)
124. Binary Tree Maximum Path Sum https://www.cnblogs.com/grandyang/p/4280120.html 如果你要计算加上当前节点的最大pa ...
- LeetCode: Binary Tree Maximum Path Sum 解题报告
Binary Tree Maximum Path SumGiven a binary tree, find the maximum path sum. The path may start and e ...
- 【LeetCode】124. Binary Tree Maximum Path Sum
Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and ...
- 二叉树系列 - 二叉树里的最长路径 例 [LeetCode] Binary Tree Maximum Path Sum
题目: Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start ...
- 第四周 Leetcode 124. Binary Tree Maximum Path Sum (HARD)
124. Binary Tree Maximum Path Sum 题意:给定一个二叉树,每个节点有一个权值,寻找任意一个路径,使得权值和最大,只需返回权值和. 思路:对于每一个节点 首先考虑以这个节 ...
- [LeetCode] Binary Tree Maximum Path Sum 求二叉树的最大路径和
Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. ...
- LeetCode(124) Binary Tree Maximum Path Sum
题目 Given a binary tree, find the maximum path sum. For this problem, a path is defined as any sequen ...
随机推荐
- while小问题
while(!m_SMque.pop(data)); 看到这个有点忘了,如果pop返回false会一直执行pop,其实这个执行的是空语句,而while每次执行都需要判断条件,所以如果pop返回fals ...
- 连接ACCESS 数据库不能使用 '';文件已在使用中。
错误类型: Microsoft JET Database Engine (0x80004005) 不能使用 '':文件已在使用中. 对数据库的操作完之后,要 conn.close() 错误原因:解 ...
- javascript陷阱,一不小心你就中招了
- 使用 Microsoft Word 发布博客文章
以 Microsoft Word 2010 为例: 依次选择:文件 -> 保存并发送 -> 发布为博客文章 配置说明:新建账户 的 博客文章 URL 一栏填写 http://rpc.cn ...
- Why we have to use epsg:900913 in OpenLayers
reference:http://docs.openlayers.org/library/spherical_mercator.html epsg:900913 is spicfy the Soher ...
- codeforces 442C C. Artem and Array(有深度的模拟)
题目 感谢JLGG的指导! 思路: //把数据转换成一条折线,发现有凸有凹 //有凹点,去掉并加上两边的最小值//无凹点,直接加上前(n-2)个的和(升序)//数据太大,要64位//判断凹与否,若一边 ...
- C#编程使用Managed Wifi API连接无线SSID
C#编程使用Managed Wifi API连接无线SSIDhttp://www.2cto.com/kf/201307/227623.html Managed Wifi API - Homehttp: ...
- Python之socketserver源码分析
一.socketserver简介 socketserver是一个创建服务器的框架,封装了许多功能用来处理来自客户端的请求,简化了自己写服务端代码.比如说对于基本的套接字服务器(socket-based ...
- C#中的 序列化和反序列化
什么是序列化和反序列化? 序列化就是把一个对象保存到一个文件或数据库字段中去,反序列化就是在适当的时候把这个文件再转化成原来的对象使用. 我想最主要的作用有: 1.在进程下次启动时读取上次保存的对象的 ...
- 关于在linux下清屏的几种技巧
在windows的DOS操作界面里面,清屏的命令是cls,那么在linux 里面的清屏命令是什么呢?下面笔者分享几种在linux下用过的清屏方法. 1.clear命令.这个命令将会刷新屏幕,本质上只是 ...