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, Yue

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 [-1000 1000] 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
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std; const int maxn=; struct Node
{
int data;
int layer;
Node *lchild,*rchild;
}; void insert(Node * & root,int data)
{
if(root==NULL)
{
root=new Node;
root->lchild=NULL;
root->rchild=NULL;
root->data=data;
return ;
}
if(root->data<data) insert(root->rchild,data);
else insert(root->lchild,data);
} int max_layer=;
int layer[maxn]={}; void layerOrder(Node * root)
{
queue<Node *> q;
root->layer=;
q.push(root);
while(!q.empty())
{
Node * now=q.front();
q.pop();
if(now->layer>max_layer) max_layer=now->layer;
layer[now->layer]+=;
if(now->lchild!=NULL)
{
now->lchild->layer=now->layer+;
q.push(now->lchild);
}
if(now->rchild!=NULL)
{
now->rchild->layer=now->layer+;
q.push(now->rchild);
}
}
} int main()
{
int n;
cin>>n;
Node * root=NULL;
for(int i=;i<n;i++)
{
int input;
cin>>input;
insert(root,input);
}
layerOrder(root);
int a=layer[max_layer];
int b=layer[max_layer-];
cout<<a<<" + "<<b<<" = "<<a+b<<endl;
return ;
}

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

  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 ...

  5. 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 ...

  6. 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& ...

  7. 1115 Counting Nodes in a BST (30 分)

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

  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. 利用谷歌插件破解今日头条的新闻ajax参数加密,新手都能懂

    最近在学习谷歌插件,想找个项目练练手,就拿今日头条开刀 首先访问地址是:https://www.toutiao.com/c/user/50025817786/#mid=50044041847 通过抓包 ...

  2. LFS搭建第一天

    1. 前期准备 vmware 软件安装 LFS iso 下载:http://ftp.osuosl.org/pub/lfs-livecd/lfslivecd-x86-6.3-r2145.iso 2.新建 ...

  3. SQL学习笔记:基础教程

    SQL语法 在表中选择列 select 列名 from 表名 选择所有列 select * from 表名 返回唯一值 select distinct 列名 from 表名 where select ...

  4. 《Java核心技术36讲》阅读笔记:谈谈对Java平台的理解笔记

    1. 谈谈你对Java平台的理解. Java是一种面向对象的语言,最显著的特性有两个方面: 一个就是一次编译,到处运行(Write once, run anywhere),能够非常容易的获得跨平台能力 ...

  5. 20155310马英林 实验2 Windows口令破解

    实 验 报 告 实验名称: 实验二 口令破解 姓名:马英林 学号: 20155310 班级: 1553 日期: 2017.10.24 一. 实验环境 •系统环境:Windows •网络环境:交换网络结 ...

  6. 20155319 《Java程序设计》实验五(网络编程与安全)实验报告

    20155319 <Java程序设计>实验五(网络编程与安全)实验报告 一.实验内容及步骤 (一) 两人一组结对编程 参考http://www.cnblogs.com/rocedu/p/6 ...

  7. WPF DataGrid使用简介

    1)自动生成列 <DataGrid AutoGenerateColumns="True" Name="datagrid" CanUserAddRows=& ...

  8. 【BZOJ2754】[SCOI2012]喵星球上的点名

    [BZOJ2754][SCOI2012]喵星球上的点名 题面 bzoj 洛谷 题解 这题有各种神仙做法啊,什么暴力\(AC\)自动机.\(SAM\)等等五花八门 我这个蒟蒻在这里提供一种复杂度正确且常 ...

  9. 1109: [POI2007]堆积木Klo

    1109: [POI2007]堆积木Klo https://lydsy.com/JudgeOnline/problem.php?id=1109 分析: 首先是dp,f[i]表示到第i个的最优值,f[i ...

  10. Lite OS学习之事件EVENT

    1. Lite OS的事件EVENT,就是一个任务向另外一个任务通知事件的,不能数据传输.看下有的函数,实际比较复杂 2. 具体还是看编程,先全局结构体整个事件变量 /*事件控制结构体*/ EVENT ...