leetcode — binary-tree-maximum-path-sum
/**
*
* Source : https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/
*
*
* Given a binary tree, find the maximum path sum.
*
* The path may start and end at any node in the tree.
*
* For example:
* Given the below binary tree,
*
* 1
* / \
* 2 3
*
* Return 6.
*/
public class BinaryTreeMaximumPathSum {
/**
* 求出遍历树节点的时候最大和,
* 可以从任何地方开始遍历,可以在任何地方结束
*
* 分析得出有两种遍历方式:
* 1. 从根节点到叶子节点之间的一段,root-leaf path中的一段
* 2. 两个节点之间经过最小公共祖先节点的path
*
* 分别对于两种情况求出最大值,然后比较求出较大的一个
*
* 定义:
* sum1为第一种情况下,一个节点可能出现的最大值
* sum1(root) = max(max(sum1(root.left), 0), max(sum1(root.right), 0)) + root.value
*
* sum2为第二种情况下,一个节点可能出现的最大值
* sum2(root) = max(sum1(root.left), 0) + max(sum1(root.right), 0) + root.value
*
*
* @param root
* @return
*/
public int maxPathSum (TreeNode root) {
MaxSumHolder holder = new MaxSumHolder();
holder.value = Integer.MIN_VALUE;
return recursion(root, holder);
}
public int recursion (TreeNode root, MaxSumHolder holder) {
if (root == null) {
return 0;
}
int sum1Left = 0;
int sum1Right = 0;
if (root.leftChild != null) {
sum1Left = Math.max(recursion(root.leftChild, holder), 0);
}
if (root.rightChild != null) {
sum1Right = Math.max(recursion(root.rightChild, holder), 0);
}
int sum1 = Math.max(sum1Left, sum1Right) + root.value;
int sum2 = sum1Left + sum1Right + root.value;
holder.value = Math.max(holder.value, Math.max(sum1, sum2));
return holder.value;
}
private class MaxSumHolder {
int value;
}
public TreeNode createTree (char[] treeArr) {
TreeNode[] tree = new TreeNode[treeArr.length];
for (int i = 0; i < treeArr.length; i++) {
if (treeArr[i] == '#') {
tree[i] = null;
continue;
}
tree[i] = new TreeNode(treeArr[i]-'0');
}
int pos = 0;
for (int i = 0; i < treeArr.length && pos < treeArr.length-1; i++) {
if (tree[i] != null) {
tree[i].leftChild = tree[++pos];
if (pos < treeArr.length-1) {
tree[i].rightChild = tree[++pos];
}
}
}
return tree[0];
}
private class TreeNode {
TreeNode leftChild;
TreeNode rightChild;
int value;
public TreeNode(int value) {
this.value = value;
}
public TreeNode() {
}
}
public static void main(String[] args) {
BinaryTreeMaximumPathSum maximumPathSum = new BinaryTreeMaximumPathSum();
char[] arr = new char[]{'1','2','3'};
System.out.println( maximumPathSum.maxPathSum(maximumPathSum.createTree(arr)) + "----6");
}
}
leetcode — binary-tree-maximum-path-sum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 解题报告
Binary Tree Maximum Path SumGiven a binary tree, find the maximum path sum. The path may start and e ...
- 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 ...
- [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–Binary Tree Maximum Path Sum
1.题目说明 Given a binary tree, find the maximum path sum. The path may start and end at any node in t ...
- C++ leetcode Binary Tree Maximum Path Sum
偶然在面试题里面看到这个题所以就在Leetcode上找了一下,不过Leetcode上的比较简单一点. 题目: Given a binary tree, find the maximum path su ...
- [LeetCode] Binary Tree Maximum Path Sum(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]Binary Tree Maximum Path Sum @ Python
原题地址:https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/ 题意: Given a binary tree, find th ...
- [Leetcode] Binary tree maximum path sum求二叉树最大路径和
Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. ...
- LeetCode Binary Tree Maximum Path Sum 二叉树最大路径和(DFS)
题意:给一棵二叉树,要求找出任意两个节点(也可以只是一个点)的最大路径和,至少1个节点,返回路径和.(点权有负的.) 思路:DFS解决,返回值是,经过从某后代节点上来到当前节点且路径和最大的值.要注意 ...
随机推荐
- JS浏览器兼容问题
一.JS与DOM的兼容性: (一) DOM节点的访问: 1.以前对DOM节点访问一般用“document.All.元素ID属性值”或者“document.元素ID属性值”这种简化的方法,在FireFo ...
- LeetCode 31 Next Permutation / 60 Permutation Sequence [Permutation]
LeetCode 31 Next Permutation / 60 Permutation Sequence [Permutation] <c++> LeetCode 31 Next Pe ...
- hadoop源码学习(二)之ZooKeeper
要能够熟练使用hadoop,就得对其原理和源码有些了解.hadoop中比较重要的概念是NameNode,DataNode,去看这些类时,又会发现其使用了ZooKeeper包,这样就可以将hadoop的 ...
- 32 ArcToolBox学习系列之数据管理工具箱——属性域(Domains)的两种创建及使用方式
属性域分为两类,一种是范围域,一种是编码的值,下面将两个一起介绍,其中涉及到的编码,名称,只是试验,并非真实情况. 一.首先新建一个文件型地理数据库,将数据导入或者是新建要素类都可以 二.打开ArcT ...
- Jmeter中连接Oracle报错Cannot create PoolableConnectionFactory
填坑贴,之前一直用jmeter2.13版本进行oracle测试,今天改为3.2版本,发现按照以往的方法执行测试,JDBC Request结果始终报错:Cannot create PoolableCon ...
- Hibernate4集成spring4报错----No Session found for current thread
在编写一个Hibernate4集成spring4的小demo的时候出现了该错误: org.hibernate.HibernateException: No Session found for curr ...
- Mesos源码分析(16): mesos-docker-executor的运行
mesos-docker-executor的运行代码在src/docker/executor.cpp中 int main(int argc, char** argv) { GOOGLE_PRO ...
- 琐事集 vol 2
vol 2-0 宝宝,你是不是该看书咯? 她正瘫在沙发上看剧 我刚提起看书,她惊恐地看了看我 然后眼白一翻,彻底地瘫平了 宝宝? “宝宝睡着了.” 你下周就要考护师了!很难得,她认真地睁开眼,信誓旦旦 ...
- [.net 面向对象程序设计深入](26)实战设计模式——策略模式 Strategy (行为型)
[.net 面向对象程序设计深入](26)实战设计模式——策略模式 Strategy (行为型) 1,策略模式定义 策略模式定义了一系列的算法,并将每一个算法封装起来,而且使它们还可以相互替换.策略模 ...
- 中文转拼音without CJK
Xamarin写Android程序时,通常要使用按中文首字母分组显示(如通讯录) . 于是需要被迫包含CJK,不过包含后包肯定是会变大的,于是....自己写了一个硬枚举的中文转拼音的类. 原理是这样的 ...