1115 Counting Nodes in a BST (30 分)

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

  • The left subtree of a node contains only nodes with keys less than or equal to the node's key.
  • The right subtree of a node contains only nodes with keys greater than the node's key.
  • Both the left and right subtrees must also be binary search trees.

Insert a sequence of numbers into an initially empty binary search tree. Then you are supposed to count the total number of nodes in the lowest 2 levels of the resulting tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤1000) which is the size of the input sequence. Then given in the next line are the N integers in [−10001000] which are supposed to be inserted into an initially empty binary search tree.

Output Specification:

For each case, print in one line the numbers of nodes in the lowest 2 levels of the resulting tree in the format:

n1 + n2 = n

where n1 is the number of nodes in the lowest level, n2 is that of the level above, and n is the sum.

Sample Input:

9
25 30 42 16 20 20 35 -5 28

Sample Output:

2 + 4 = 6

题意:求二叉查找树最低两层结点,并求出他们的和。

分析:先建树,再层序遍历求出最大层数maxlayer,遍历过程中用num数组来记录每层结点个数即可。还有一种方法是用DFS求出最大深度maxdepth,同样也用数组记录每个深度的结点个数。

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-27-15.33.36
 * Description : A1115
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 #include<queue>
 using namespace std;
 ;
 int a[maxn],n;
 int num[maxn];

 struct node{
     int data,layer;
     node* lchild;
     node* rchild;
 };

 node* newNode(int v){
     node* Node=new node;
     Node->data=v;
     Node->lchild=Node->rchild=NULL;
     return Node;
 }
 void insert(node* &root,int x){
     if(root==NULL) {
         root=newNode(x);
         return;
     }
     else if(x>root->data) insert(root->rchild,x);
     else insert(root->lchild,x);
 }
 node* Create(int data[],int n){
     node* root=NULL;
     ;i<n;i++){
         insert(root,data[i]);
     }
     return root;
 }
 ;
 void BFS(node* root){
     queue<node*> q;
     q.push(root);
     root->layer=;
     while(!q.empty()){
         node* now=q.front();
         q.pop();
         num[now->layer]++;
         if(maxlayer<now->layer) maxlayer=now->layer;
         if(now->lchild){
             now->lchild->layer=now->layer+;
             q.push(now->lchild);
         }
         if(now->rchild){
             now->rchild->layer=now->layer+;
             q.push(now->rchild);
         }
     }
 }

 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     scanf("%d",&n);
     ;i<n;i++){
         scanf("%d",&a[i]);
     }
     node* root=Create(a,n);
     BFS(root);
     printf(],num[maxlayer]+num[maxlayer-]);

     ;
 }
 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-06-00.18.21
 * Description : A1115
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 using namespace std;
 ;
 struct node{
     int data;
     node* lchild;
     node* rchild;
 };

 node* newNode(int v){
     node* Node=new node;
     Node->data=v;
     Node->lchild=Node->rchild=NULL;
     return Node;
 }
 void insert(node* &root,int x){
     if(root==NULL) {
         root=newNode(x);
         return;
     }
     else if(x>root->data) insert(root->rchild,x);
     else insert(root->lchild,x);
 }
 node* Create(int data[],int n){
     node* root=NULL;
     ;i<n;i++){
         insert(root,data[i]);
     }
     return root;
 }
 },maxdepth=-;
 void DFS(node* root,int depth){
     if(root==NULL){
         maxdepth=max(maxdepth,depth);
         return;
     }
     num[depth]++;
     DFS(root->lchild,depth+);
     DFS(root->rchild,depth+);
 }
 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     int n,data[maxn];
     cin>>n;
     ;i<n;i++){
         cin>>data[i];
     }
     node* root=Create(data,n);
     DFS(root,);
     printf(],num[maxdepth-],num[maxdepth-]+num[maxdepth-]);

     ;
 }

1115 Counting Nodes in a BST (30 分)的更多相关文章

  1. 【PAT甲级】1115 Counting Nodes in a BST (30分)(二叉查找树)

    题意: 输入一个正整数N(<=1000),接着输入N个整数([-1000,1000]),依次插入一棵初始为空的二叉排序树.输出最底层和最底层上一层的结点个数之和,例如x+y=x+y. AAAAA ...

  2. [二叉查找树] 1115. Counting Nodes in a BST (30)

    1115. Counting Nodes in a BST (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Y ...

  3. 1115. Counting Nodes in a BST (30)

    A Binary Search Tree (BST) is recursively defined as a binary tree which has the following propertie ...

  4. PAT Advanced 1115 Counting Nodes in a BST (30) [⼆叉树的遍历,BFS,DFS]

    题目 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following proper ...

  5. PAT甲题题解-1115. Counting Nodes in a BST (30)-(构建二分搜索树+dfs)

    题意:给出一个序列,构建二叉搜索树(BST),输出二叉搜索树最后两层的节点个数n1和n2,以及他们的和sum: n1 + n2 = sum 递归建树,然后再dfs求出最大层数,接着再dfs计算出最后两 ...

  6. PAT A 1115. Counting Nodes in a BST (30)【二叉排序树】

    题目:二叉排序树,统计最后两层节点个数 思路:数组格式存储,insert建树,dfs遍历 #include<cstdio> #include<iostream> #includ ...

  7. PAT (Advanced Level) 1115. Counting Nodes in a BST (30)

    简单题.统计一下即可. #include<cstdio> #include<cstring> #include<cmath> #include<vector& ...

  8. PAT甲1115 Counting Nodes in a BST【dfs】

    1115 Counting Nodes in a BST (30 分) A Binary Search Tree (BST) is recursively defined as a binary tr ...

  9. PAT 1115 Counting Nodes in a BST[构建BST]

    1115 Counting Nodes in a BST(30 分) A Binary Search Tree (BST) is recursively defined as a binary tre ...

随机推荐

  1. VS2012调用64位IIS Express

    在注册表键HKEY_CURRENT_USER\Software\Microsoft\VisualStudio\11.0\WebProjects下添加DWORD "Use64BitIISExp ...

  2. Java平台标准版本

    JDK   Java Language Java Language     Tools &Tool APIs java javac javadoc jar javap jdeps Script ...

  3. 2016 ACM/ICPC Asia Regional Qingdao Online 1001 I Count Two Three(打表+二分搜索)

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission( ...

  4. js 压缩 预览 上传图片

    com.js export const compressImage=function (files,fn,preCallbackFn,i) { let newfile = files.files[0] ...

  5. (4)socket的基础使用(基于TCP协议的并发编程)

    需要实现并发需要依靠socketserver 模块 socketserver模块下有几个功能 def __init__(self, request, client_address, server): ...

  6. hello1与hello2的代码分析

    1.hello1代码分析 hello.java package javaeetutorial.hello1; import javax.enterprise.context.RequestScoped ...

  7. 【java规则引擎】《Drools7.0.0.Final规则引擎教程》第4章 4.2 activation-group& dialect& date-effective

    转载至:https://blog.csdn.net/wo541075754/article/details/75511887 activation-group 该属性将若干个规则划分成一个组,统一命名 ...

  8. 我的nginx iis 负载均衡学习(环境搭建)

    1,下载并安装nginx 比较简单 2,进行网站的配置 我使用了我的IIS 站点中已经拥有的两个站点 3,进行nginx 的配置 配置如下: 在server 节点之前添加如下的配置: upstream ...

  9. No result defined for action com.nynt.action.ManageAction and result input问题

    No result defined for action com.nynt.action.ManageAction and result input 问题原因: 1). 在action类中定义的一个r ...

  10. 【MVC】Model的使用

    1,Model的职责: Model只负责与数据处理相关的工作. 2,开发Model的基本观念 采用ORM信息访问技术开发 ORM是将结构化的关系型数据,映射成面向对象模型.对于EF来说,就是关系型数据 ...