【二叉树的递归】05二叉树中找任意起点和终点使他们的路径和最大【Binary Tree Maximum Path Sum】

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

给定一个二叉树，寻找值最大的路径。

这个路径可以从这个树上面的任意一个节点开始，然后在任意一个节点结束。

例如：

给定下面的二叉树，

返回 6.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

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,

Return 6.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

test.cpp:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77

#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include "BinaryTree.h"

using namespace std;

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
int max_sum;
int dfs(const TreeNode *root)
{
    if(root == NULL)
    {
        return 0;
    }
    int l = dfs(root->left);
    int r = dfs(root->right);
    int sum = root->val;
    if(l > 0)
    {
        sum += l;
    }
    if(r > 0)
    {
        sum += r;
    }
    max_sum = max(max_sum, sum);
    return max(r, l) > 0 ? max(r, l) + root->val : root->val;
}
int maxPathSum(TreeNode *root)
{

max_sum = -0xffff;
dfs(root);
return max_sum;
}

// 树中结点含有分叉，
//                  8
//              /       \
//             6         1
//           /   \
//          9     2
//               / \
//              4   7
int main()
{
    TreeNode *pNodeA1 = CreateBinaryTreeNode(8);
    TreeNode *pNodeA2 = CreateBinaryTreeNode(6);
    TreeNode *pNodeA3 = CreateBinaryTreeNode(1);
    TreeNode *pNodeA4 = CreateBinaryTreeNode(9);
    TreeNode *pNodeA5 = CreateBinaryTreeNode(2);
    TreeNode *pNodeA6 = CreateBinaryTreeNode(4);
    TreeNode *pNodeA7 = CreateBinaryTreeNode(7);

ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);

PrintTree(pNodeA1);

cout << maxPathSum(pNodeA1) << endl;

DestroyTree(pNodeA1);
return 0;
}

结果输出：

BinaryTree.h:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode *CreateBinaryTreeNode(int value);
void ConnectTreeNodes(TreeNode *pParent,
TreeNode *pLeft, TreeNode *pRight);
void PrintTreeNode(TreeNode *pNode);
void PrintTree(TreeNode *pRoot);
void DestroyTree(TreeNode *pRoot);

#endif /*_BINARY_TREE_H_*/

BinaryTree.cpp:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89

#include <iostream>
#include <cstdio>
#include "BinaryTree.h"

using namespace std;

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

//创建结点
TreeNode *CreateBinaryTreeNode(int value)
{
TreeNode *pNode = new TreeNode(value);

return pNode;
}

//连接结点
void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight)
{
    if(pParent != NULL)
    {
        pParent->left = pLeft;
        pParent->right = pRight;
    }
}

//打印节点内容以及左右子结点内容
void PrintTreeNode(TreeNode *pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->val);

if(pNode->left != NULL)
            printf("value of its left child is: %d.\n", pNode->left->val);
        else
            printf("left child is null.\n");

if(pNode->right != NULL)
            printf("value of its right child is: %d.\n", pNode->right->val);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }

printf("\n");
}

//前序遍历递归方法打印结点内容
void PrintTree(TreeNode *pRoot)
{
PrintTreeNode(pRoot);

if(pRoot != NULL)
    {
        if(pRoot->left != NULL)
            PrintTree(pRoot->left);

if(pRoot->right != NULL)
PrintTree(pRoot->right);
}
}

void DestroyTree(TreeNode *pRoot)
{
    if(pRoot != NULL)
    {
        TreeNode *pLeft = pRoot->left;
        TreeNode *pRight = pRoot->right;

delete pRoot;
pRoot = NULL;

DestroyTree(pLeft);
DestroyTree(pRight);
}
}

【二叉树的递归】05二叉树中找任意起点和终点使他们的路径和最大【Binary Tree Maximum Path Sum】的更多相关文章

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 124. 二叉树中的最大路径和（Binary Tree Maximum Path Sum）
题目描述给定一个非空二叉树,返回其最大路径和. 本题中,路径被定义为一条从树中任意节点出发,达到任意节点的序列.该路径至少包含一个节点,且不一定经过根节点. 示例 1: 输入: [1,2,3] 1 ...
LeetCode 124. Binary Tree Maximum Path Sum 二叉树中的最大路径和 (C++/Java)
题目: Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as ...
[Swift]LeetCode124. 二叉树中的最大路径和 | Binary Tree Maximum Path Sum
Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as any ...
[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 求二叉树的最大路径和
Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as any ...
75.Binary Tree Maximum Path Sum（二叉树的最大路径和）
Level: Hard 题目描述: Given a non-empty binary tree, find the maximum path sum. For this problem, a pa ...
[leetcode]124. Binary Tree Maximum Path Sum二叉树最大路径和
Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as any ...
[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. ...

随机推荐

信号量semaphore解析
1 基础概念信号量在创建时须要设置一个初始值,表示同一时候能够有几个任务能够訪问该信号量保护的共享资源.初始值为1就变成相互排斥锁(Mutex),即同一时候仅仅能有一个任务能够訪问信号量保护的共享资 ...
Swift_4_闭包(Blocks)
import Foundation println("Hello, World!") var arr = [1,2,4,6,74,2] func hasClosure(list:[ ...
iOS表格制作
由于项目上的需求,需要做一个表格出来,来显示流程状态.刚开始脑子一头雾水,没有一点思路,但是靠着自己的座右铭--“世上无难事,只怕有心人”,克服了所有困难.好,不说了,讲正事. 制作表格,还是需要ta ...
【BZOJ3510】首都 LCT维护子树信息+启发式合并
[BZOJ3510]首都 Description 在X星球上有N个国家,每个国家占据着X星球的一座城市.由于国家之间是敌对关系,所以不同国家的两个城市是不会有公路相连的. X星球上战乱频发,如果A国打 ...
zabbix server is not running解决办法
正常安装完zabbix后,登录后zabbix监控报错zabbix server is not running: the information displayed may not be current ...
我的Android进阶之旅------> Android在TextView中显示图片方法
面试题:请说出Android SDK支持哪些方式显示富文本信息(不同颜色.大小.并包含图像的文本信息),并简要说明实现方法. 答案:Android SDK支持如下显示富文本信息的方式. 1.使用Tex ...
聊聊数据库~5.SQL运维上篇
1.6.SQL运维篇运维这块逆天只能说够用,并不能说擅长,所以这篇就当抛砖之用,欢迎补充和纠错 PS:再说明下CentOS优化策略这部分的内容来源:首先这块逆天不是很擅长,所以主要是参考网上的DBA ...
11.23 Eclipse
一.Eclipse Workspace File-->Switch Workspace 编译环境:Window -- Preferences – Java - Compiler 运行环境:Win ...
HDU - 1175 连连看【DFS】【BFS】
题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1175 思路这种题一想到就用搜索, 但是内存是32m 用 bfs 会不会MLE 没错第一次 BFS的 ...
面向对象分析与设计（C++）课堂笔记
第一次课: 对象是程序设计最基本的单元对象:对象标识.属性.操作(对象标识又分为内部标识.外部标识) 三三制原则继承:英文语义”is a kind of” 自动的拥有或隐含的复制虚基类:解决多继 ...

【二叉树的递归】05二叉树中找任意起点和终点使他们的路径和最大【Binary Tree Maximum Path Sum】

【二叉树的递归】05二叉树中找任意起点和终点使他们的路径和最大【Binary Tree Maximum Path Sum】的更多相关文章

随机推荐

热门专题