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

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,

       1
/ \
2 3

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,

       1
/ \
2 3

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:
24
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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. [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 ...

  5. [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. ...

  6. [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 ...

  7. 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 ...

  8. [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 ...

  9. [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

  1. textarea中的内容的获取

    今天他们说为啥获取不到textarea的数值 这个问题让我很纳闷  为什么会获取不到呢? 按照逻辑来说 同样都是表单元素  怎么可能出现呢? 我就看了一眼代码 alert($("#texta ...

  2. postman是如何使用的?

    1.地址:https://segmentfault.com/a/1190000005055899

  3. NSNotificationCenter详解

    本文转载至 http://blog.csdn.net/chengyingzhilian/article/details/7874408 作用:NSNotificationCenter是专门供程序中不同 ...

  4. SQLServer2008 统计表占用空间

    1.查看一张表占用的空间大小 EXEC sp_spaceused '表' 1.查看一个数据库中所有表占用空间大小 EXEC sp_MSforeachtable 'sp_spaceused " ...

  5. MATLAB循环结构:while语句P69范数待编

    while语句的一般格式为: while 条件 循环体语句 end 从键盘输入若干个数,当输入0时结束输入,求这些数的平均值和它们的和. 程序如下: sum=; n=; x=input('输入一个数字 ...

  6. java服务安装(一):使用java service wrapper及maven打zip包

    目录(?)[+] 1概述 1_1为什么要用服务形式运行 1_2如何让java程序以服务形式运行 1_3打包需求 2程序示例 3maven打zip包 3_1maven-assembly-plugin介绍 ...

  7. Intellij IDEA打开多项目窗口

    我版本是2016.02.04 其他版本可能不一样的设置

  8. 【基于rssi室内定位报告】rssi分布情况标识位置

    import matplotlib matplotlib.use('Agg') import numpy as np from numpy import array from matplotlib i ...

  9. Android笔记之获取debug.keystore和release.keystore的MD5/SHA1值

    获取debug.keystore的key,如下图 获取release.keystore的key 输入命令keytool -list -v -keystore <jksFilename> 例 ...

  10. 我的Android进阶之旅------>/storage/sdcard0, /sdcard, /mnt/sdcard ,/storage/emulated/legacy 的区别

    转自:http://bbs.gfan.com/android-5382920-1-1.html 关于android的4.2的0文件夹的详解---- android 4.0 ----在galaxy ne ...