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

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. Redis加锁与解锁

    Redis加锁 customerM = BaseMemCached.setMLock(customerId); /** * 个人账户表加锁 **/ public static CustomerM se ...

  2. 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 ...

  3. 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 ...

  4. windy数(简单数位DP)

    1026: [SCOI2009]windy数 Time Limit: 1 Sec  Memory Limit: 162 MBSubmit: 6306  Solved: 2810[Submit][Sta ...

  5. 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 ...

  6. 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 ...

  7. centos 安装 jdk PostgreSQL

    1.下载: anzhuang  jDK http://blog.csdn.net/youzhouliu/article/details/51183115 ----------------------- ...

  8. 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 ...

  9. 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 ...

  10. 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', ...