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
- javascript高级函数
高级函数 安全的类型检测 js内置的类型检测并非完全可靠,typeof操作符难以判断某个值是否为函数 instanceof在多个frame的情况下,会出现问题. 例如:var isArray = va ...
- 为了让vi命令也可以使用vim的配置,需要修改 vi /etc/bashrc 增加一行 alias vi='vim'此时,经过上面配置已经可以显示语法高亮了
为了让vi命令也可以使用vim的配置,需要修改 vi /etc/bashrc 增加一行 aliasvi='vim'此时,经过上面配置已经可以显示语法高亮了
- WPF编程学习——样式
本文目录 1.引言 2.怎样使用样式? 3.内联样式 4.已命名样式 5.元素类型样式 6.编程控制样式 7.触发器 1.引言 样式(Style),主要是用来让元素或内容呈现一定外观的属性.WPF中的 ...
- RT/Metro商店应用如何如何获取图片的宽高
RT/Metro商店应用如何如何获取图片的宽高 var file = await StorageFile.GetFileFromApplicationUriAsync(new Uri("ms ...
- 表单中<form>的enctype属性
application/x-www-form-urlencoded.multipart/form-data.text/plain 上传文件的表单中<form>要加属性enctype=&qu ...
- Oracle自带的用户
Oracle安装完毕创建数据库实例的时候,会自动生成三个用户sys,system,scott. sys用户是超级管理员,具有最高权限,充当sysdba角色,可以执行create database,默认 ...
- Tech Stuff - Mobile Browser ID (User-Agent) Strings
Tech Stuff - Mobile Browser ID (User-Agent) Strings The non-mobile stuff is here (hint: you get jerk ...
- 全国DNS汇总
全国DNS汇总 来路不明的DNS服务器可能导致你的帐号密码轻易被盗,请谨慎使用!在中国大陆,最科学的方法是将首选DNS服务器设置为114.114.114.114,备用DNS服务器设置为当地电信运营商的 ...
- 饶有兴致的用javascript做了个贪食蛇游戏
09年写的东西.一直藏在自己的记事本里头,现在开始整理写博客,所以直接搬过来 先上效果图 再添代码: <HTML> <HEAD> <TITLE>贪吃蛇 Snake ...
- Nutch配置:nutch-default.xml详解
/×××××××××××××××××××××××××××××××××××××××××/ Author:xxx0624 HomePage:http://www.cnblogs.com/xxx0624/ ...