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

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. Django安装和启动

    1.django安装 在http://www.djangoproject.com/download/这个网站上可以下载django的最新版本.在下载时,要注意django版本和本机安装的Python版 ...

  2. 承载(Host)通用语言执行时

    承载(Host)通用语言执行时(CLR) 还有一种使用COM 的方法是是把须要集成的 F# 代码与已有的 C/C++ 应用程序集成到一起.开成自己定义的承载通用语言执行时.通用语言执行时就是 C++ ...

  3. POJ 2187 Beauty Contest【旋转卡壳求凸包直径】

    链接: http://poj.org/problem?id=2187 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#probl ...

  4. SQLServer判断一个IP是否在一个IP段里

    declare @ip1 varchar(20)declare @ip2 varchar(20)set @ip1='221.231.138.101'set @ip2='255.255.255.255' ...

  5. 我的Android进阶之旅------>Android中AsyncTask源码分析

    在我的<我的Android进阶之旅------>android异步加载图片显示,并且对图片进行缓存实例>文章中,先后使用了Handler和AsyncTask两种方式实现异步任务机制. ...

  6. Open AI Gym简介

    介绍 OpenAI Gym是一款用于研发和比较强化学习算法的工具包,它支持训练智能体(agent)做任何事——从行走到玩Pong或围棋之类的游戏都在范围中. OpenAI Gym 是一个用于开发和比较 ...

  7. java面向对象入门之方法参数的传递

    /* Name : Power by :Stuart Date:2015.4.25 */ class PassOn{ //创建show方法,把i传入,输出i+1的结果 public void show ...

  8. Android app与PC端交互

    app提交信息到PC端mysql数据库 新建名为SignActivity package com.example.administrator.success; import android.app.A ...

  9. mysql主从复制(windows下)

    简单搭建mysql主从服务器(双向复制),仅供学习之用. 1.下载mysql-advanced-5.6.16-winx64.zip 下载完成后解压,重命名成mysql,分别放入两台服务器(看第二点集群 ...

  10. PAT 天梯赛 L2-009. 抢红包 【排序】

    题目链接 https://www.patest.cn/contests/gplt/L2-009 思路 用结构体存储,然后结构体排序 注意一下 个人编号是从 1 开始 计数的 AC代码 #include ...