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

  1. [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 ...

  2. 【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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 【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 ...

  7. 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 ...

  8. 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 ...

  9. [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. ...

  10. 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

  1. JavaScript之isNaN()函数讲解

    定义和用法 isNaN() 函数用于检查其参数是否是非数字值. 语法 isNaN(x) 参数 描述 x 必需.要检测的值. 返回值 如果 x 是特殊的非数字值 NaN(或者能被转换为这样的值),返回的 ...

  2. INSERT IGNORE 与INSERT INTO的区别

      INSERT IGNORE 与INSERT INTO的区别就是INSERT IGNORE会忽略数据库中已经存在 的数据,如果数据库没有数据,就插入新的数据,如果有数据的话就跳过这条数据.这样就可以 ...

  3. Javascript引擎单线程机制及setTimeout执行原理说明

    setTimeout用法在实际项目中还是会时常遇到.比如浏览器会聪明的等到一个函数堆栈结束后才改变DOM,如果再这个函数堆栈中把页面背景先从白色设为红色,再设回白色,那么浏览器会认为DOM没有发生任何 ...

  4. 输入格式--InputFormat和InputSplit

    1)InputFormat的类图: InputFormat 直接子类有三个:DBInputFormat.DelegatingInputFormat和FileInputFormat,分别表示输入文件的来 ...

  5. 小圣求职记A:腾讯篇

    本人普通985高校计算机专业研究生一枚,从9月12号开始正式找工作,一个月过去了,参加了能参加的各个互联网公司的宣讲.笔试.面试,现用两篇随笔分享所见所闻.随笔A将以腾讯为例详细展示整个过程,随笔B将 ...

  6. 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 ...

  7. 使用jmeter对websocket进行压力测试[转载]

    前段时间本着练习angularJS+requireJS的目的写了一个基于nodeJS和socket.io的聊天室,github地址为:https://github.com/towersxu/node- ...

  8. appserv安装

    Appserv  官网: http://www.appservnetwork.com/ 安装好后,输入http://localhost:8082/验证是否装成功,成功后如下图

  9. Java:Comparator接口

    public interface Comparator<T> 接口里面的方法 int compare(T o1, T o2) o1 > o2 返回 1 o1 = o2 返回 0 o1 ...

  10. libprotobuf ERROR

    google/protobuf/wire_format.cc:1059] Encountered string containing invalid UTF-8 data while parsing  ...