【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

Redis加锁与解锁
Redis加锁 customerM = BaseMemCached.setMLock(customerId); /** * 个人账户表加锁 **/ public static CustomerM se ...
C# 关于类型转换面试题
分别分析一下两种写法是否正确.假设不对.请说明原因写法一: short s=1; s = s + 1; 写法二: short s=1; s += 1; 解答: 写法一不对,会报出以下的错误: 无法将 ...
eclipse tasks
tasks可以在代码里增加标识,通过tasks view可以快速的找到这些标识的地方,有助于提高开发效率和代码管理. 通过Eclipse的 Window==>Show View==>Tas ...
windy数(简单数位DP)
1026: [SCOI2009]windy数 Time Limit: 1 Sec Memory Limit: 162 MBSubmit: 6306 Solved: 2810[Submit][Sta ...
Java中List.remove报UnsupportedOperationException异常
今天项目中有个需求场景: A和B都是List,而B是A的子集,现在想求A和B的差集. 想到了List中提供的removeAll()方法可以求得差集,但是结果确报了UnsupportedOperatio ...
fkwの题目(祝松松生日快乐！)
麓山国际实验学校傅少,匡哥和巨夫出的题目(共3道) 一.题目概况题目名称打地铺泡妹子开房间题目类型传统传统传统可执行文件名 deeeep soccer room 输入文件名 dee ...
centos 安装 jdk PostgreSQL
1.下载: anzhuang jDK http://blog.csdn.net/youzhouliu/article/details/51183115 ----------------------- ...
遇到IIS configuration error错误的可以看看，不一定是权限问题
最近接手了别人的一个 DOT NET项目,编译.调试一切都OK(心里暗暗高兴),发布吧,结果放到服务器上一运行出现Configuration Error错误,提示:“Access to the pat ...
activiti基础--1------------------------生成.bpmn和.png以及部署流程定义
helloworld.dbmn <?xml version="1.0" encoding="UTF-8"?> <definitions xml ...
python基础5 ---python文件处理
python文件处理一.文件处理的流程打开文件,得到文件句柄并赋值给一个变量通过句柄对文件进行操作关闭文件二.文件的操作方法 1.文件打开模式格式: 文件句柄 = open('文件路径', ...

【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