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

给定一个二叉树,找出他的最小的深度。

最小的深度,指的是从根节点到叶子节点的,经历的最多的节点个数。

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

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

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

using namespace std;

int maxDepth(TreeNode *root)
{
    if(root == NULL)
    {
        return 0;
    }
    /*用队列来模拟广度优先搜索*/
    queue<TreeNode *> que;
    que.push(root);
    int count = 1;
    int depth = 0;
    while(!que.empty())
    {
        TreeNode *tmp = que.front();
        que.pop();
        count --;

if(tmp->left != NULL)
        {
            que.push(tmp->left);
        }
        if(tmp->right != NULL)
        {
            que.push(tmp->right);
        }
        if(count == 0)
        {
            depth ++;
            count = que.size();
        }
    }
    return depth;
}

// 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);

cout << maxDepth(pNodeA1) << endl;

DestroyTree(pNodeA1);
    return 0;
}

结果输出:
4
 
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】02二叉树的最大深度【Maximum Depth of Binary Tree】的更多相关文章

  1. LeetCode 104. 二叉树的最大深度(Maximum Depth of Binary Tree)

    104. 二叉树的最大深度 104. Maximum Depth of Binary Tree 题目描述 给定一个二叉树,找出其最大深度. 二叉树的深度为根节点到最远叶子节点的最长路径上的节点数. 说 ...

  2. [Swift]LeetCode104. 二叉树的最大深度 | Maximum Depth of Binary Tree

    Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the long ...

  3. [LintCode] Maximum Depth of Binary Tree 二叉树的最大深度

    Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the long ...

  4. leetcode 104 Maximum Depth of Binary Tree二叉树求深度

    Maximum Depth of Binary Tree Total Accepted: 63668 Total Submissions: 141121 My Submissions Question ...

  5. 104. Maximum Depth of Binary Tree(C++)

    104. Maximum Depth of Binary Tree Given a binary tree, find its maximum depth. The maximum depth is ...

  6. LeetCode:Minimum Depth of Binary Tree,Maximum Depth of Binary Tree

    LeetCode:Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth ...

  7. 【LeetCode练习题】Maximum Depth of Binary Tree

    Maximum Depth of Binary Tree Given a binary tree, find its maximum depth. The maximum depth is the n ...

  8. [Leetcode][JAVA] Minimum Depth of Binary Tree && Balanced Binary Tree && Maximum Depth of Binary Tree

    Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the n ...

  9. LeetCode Javascript实现 258. Add Digits 104. Maximum Depth of Binary Tree 226. Invert Binary Tree

    258. Add Digits Digit root 数根问题 /** * @param {number} num * @return {number} */ var addDigits = func ...

  10. 【LeetCode】104. Maximum Depth of Binary Tree (2 solutions)

    Maximum Depth of Binary Tree  Given a binary tree, find its maximum depth. The maximum depth is the ...

随机推荐

  1. SVN版本号控制软件-图片含义具体解释

    转载请注明出处:http://blog.csdn.net/zhuwentao2150/article/details/51195154 自己定义SVN图标显示风格 SVN的图标是能够自己定义风格的 右 ...

  2. html5小趣味知识点系列(一)spellcheck

    发现一些h5的系ode知识点增加一些趣味性 实用性 不敢妄自评论  觉得有用就用一下  没用就路过一下 spellcheck属性 它的功能是针对用户输入的文本内容进行拼写和语法检查 用于input 和 ...

  3. VirtualBox + CentOS 虚拟机网卡配置

    摘要: 要学好Linux,还是得自己搭建虚拟机. VirtualBox比较小巧简单,容易上手.在配合CentOS 6.4使用时,首要的问题就是网卡配置,尤其是使用SSH终端仿真程序(例如SecureC ...

  4. linux下apache php配置redis

    1.安装redis 第一步: 下载:https://github.com/nicolasff/phpredis/archive/2.2.4.tar.gz 上传phpredis-2.2.4.tar.gz ...

  5. Jmeter 04 JMeter 负载与监听

    1. 场景设计 2. 场景设置 3. JMeter性能参数配置 4. 测试监听

  6. P1009 阶乘之和

    P1009 阶乘之和 题目提供者洛谷OnlineJudge 标签数论(数学相关)高精1998NOIp提高组NOIp普及组 难度普及- 通过/提交1139/3791 提交该题 讨论 题解 记录 题目描述 ...

  7. visual studio code (vscode)像 sublime text 的 ctrl+d 一样多光标选中

    快捷键是 ctrl+m ,返回上一个选中时ctrl+u. 文件 ==>首选项 ==>键盘快捷键() 里面可以查到,下一个是“将选择添加到下一个查找匹配项”,返回上一个是“cursorund ...

  8. 九度OJ 1341:艾薇儿的演唱会 (最短路)

    时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:528 解决:241 题目描述: 艾薇儿今天来到了中国,她计划两天后在哈尔滨举行一场个人的演唱会.由于出现了紧急情况,演唱会的举办方要求艾薇儿 ...

  9. PBR探索

    原理 根据能量守恒,以及一系列光照原理得出微表面BRDF(Bidirectional Reflectance Distribution Function)公式 // D(h) F(v,h) G(l,v ...

  10. python cookbook第三版学习笔记十九:未包装的函数添加参数

    比如有下面如下的代码,每个函数都需要判断debug的是否为True,而默认的debug为False def a(x,debug=False): if debug: print('calling a') ...