1115. Counting Nodes in a BST (30)
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
建树,然后不断更新最大深度和次大深度,及相应点个数。
代码:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
struct tree
{
int data;
tree *left,*right;
tree()
{
data = ;
left = right = NULL;
}
}*head;
int n;
int m,c1,c2;
tree *insert_(tree *r,int d,int h)
{
if(r == NULL)
{
r = new tree();
r -> data = d;
if(h > m)
{
m = h;
c2 = c1;
c1 = ;
}
else if(h == m)c1 ++;
else if(h == m - )c2 ++;
}
else if(d > r -> data)
{
r -> right = insert_(r -> right,d,h + );
}
else
{
r -> left = insert_(r -> left,d,h + );
}
return r;
}
int main()
{
int d;
scanf("%d",&n);
head = NULL;
for(int i = ;i < n;i ++)
{
scanf("%d",&d);
head = insert_(head,d,);
}
printf("%d + %d = %d",c1,c2,c1 + c2);
}
1115. Counting Nodes in a BST (30)的更多相关文章
- [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 ...
- 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 ...
- 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 ...
- 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 ...
- 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& ...
- 【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 ...
- 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 ...
- 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 ...
- 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
- Swift开发教程--怎样设置状态栏的文字颜色
第一步:在Info.plist中设置UIViewControllerBasedStatusBarAppearance 为NO 第二步:在viewDidLoad中加一句 UIApplication.sh ...
- Python菜鸟之路:Python基础(三)
一.编码 推荐阅读<字符编码的前世今生>:http://tgideas.qq.com/webplat/info/news_version3/804/808/811/m579/201307/ ...
- Webpack探索【2】--- 安装、项目初始化、webpack.config.js配置文件
本文主要讲安装.项目初始化.webpack.config.js配置文件方面的内容.
- Django之restframework2视图三部曲
视图三部曲 下面我来来看restframework是如何将冗余的代码一步步的进行封装. 这里主要用到的是多继承 第一步mixin类编写视图 AuthorModelSerializer: class A ...
- PAT 组合数的和(15)
给定N个非0的个位数字,用其中任意2个数字都可以组合成1个2位的数字.要求所有可能组合出来的2位数字的和.例如给定2.5.8,则可以组合出:25.28.52.58.82.85,它们的和为330. 输入 ...
- centos7 Authentication failure
root@localhost ~]#su bash-4.2$ su Password: su: Authentication failure //这里切换的是系统用户,现在还不清楚为什么postgre ...
- SSH 占用数据库连接不释放问题
SSH框架的项目在訪问数据库的时候.訪问完毕后一直占用链接,不释放.导致过了一段时间后,server没挂,就是有訪问数据库的连接是时候.一直卡住 解决的方法: 1.配置spring相应的hiber ...
- Swift学习 --- 2.1基础部分
1.swift 能够省去; 2.println与print的差别就是一个能够换行一个不能够 3.swift省去了.h与.m 直接一个swift文件 4.元组能够返回多个值,元组(tuples)把多个值 ...
- 3.11课·········C#类
String类:.Length 字符的长度,返回一个int类型的值 .Trim() 去掉开头以及结尾的空格.TrimStart() 去掉字符串开头的空格.TrimEnd() 去掉字符串后面的空格 .T ...
- 第二章 python中重要的数据结构(下)
二.元组(tuple):不可变序列 跟list一样,也是一种序列,唯一不同的是,元组元素不能被修改,通常用(, ,)表示元组,也可以不加括号. #创建元组 >>> 1,2,3 (1, ...