the problem is from pat,which website is http://pat.zju.edu.cn/contests/pat-a-practise/1043

and the source code is as followed.

#include<iostream>
#include<vector>
#include<cstdio> using namespace std;
int pos; struct cell
{
cell *lchild;
cell *rchild;
int c;
cell()
{
lchild = rchild = NULL;
}
}; void insert1(cell *&root,int x)
{
if (!root)
{
root = new cell;
root->c = x;
}
else
{
if (x < root->c)
{
insert1(root->lchild,x);
}
else
insert1(root->rchild,x);
}
}
void insert2(cell *&root,int x)
{
if (!root)
{
root = new cell;
root->c = x;
}
else
if (x > root->c)
{
insert2(root->lchild, x);
}
else
insert2(root->rchild, x);
}
void preOrder(cell *root,vector<int> &v)
{
v.push_back(root->c);
if (root->lchild != NULL)
preOrder(root->lchild,v);
if (root->rchild != NULL)
preOrder(root->rchild,v);
} void postOrder(cell *root,vector<int> &v)
{
if (root->lchild != NULL)
postOrder(root->lchild, v);
if (root->rchild != NULL)
postOrder(root->rchild, v);
v.push_back(root->c);
} int main()
{
int n;
vector<int> v1,v2,v3,v;
scanf("%d",&n);
for (int i = 0; i < n; i++)
{
int x;
scanf("%d",&x);
v1.push_back(x);
}
cell *r = NULL;//THE root of bst
cell *r1 = NULL;//the root of mirror of bst;
for (int i = 0; i < n; i++)
{
insert1(r,v1[i]);
insert2(r1,v1[i]);
}
preOrder(r,v2);//the preorder of bst
preOrder(r1,v3);//the preorder of the mirror of bst if (v2 != v1 && v3 != v1)
{
printf("NO\n");
return 0;
}
else
{
printf("YES\n");
if (v2 == v1)
{
postOrder(r,v);
for (int i = 0; i < n; i++)
{
printf("%d%c",v[i], ((i - n + 1) ? ' ' : '\n'));//this type of presentation is fitted for the pattern of pat. }
}
else if(v3 == v1)
{
postOrder(r1,v);
for (int i = 0; i < n; i++)
printf("%d%c", v[i], ((i - n + 1) ? ' ' : '\n'));
}
}
return 0;
}

i think what is the most important for the bst-like problem is build a bst. that’s the point.

as long as the bst is built,then some trial things remains.

1043. Is It a Binary Search Tree (25)的更多相关文章

  1. PAT 甲级 1043 Is It a Binary Search Tree (25 分)(链表建树前序后序遍历)*不会用链表建树 *看不懂题

    1043 Is It a Binary Search Tree (25 分)   A Binary Search Tree (BST) is recursively defined as a bina ...

  2. PAT Advanced 1043 Is It a Binary Search Tree (25) [⼆叉查找树BST]

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

  3. 1043 Is It a Binary Search Tree (25分)(树的插入)

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

  4. PAT 1043 Is It a Binary Search Tree (25分) 由前序遍历得到二叉搜索树的后序遍历

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

  5. PAT (Advanced Level) 1043. Is It a Binary Search Tree (25)

    简单题.构造出二叉搜索树,然后check一下. #include<stdio.h> #include<algorithm> using namespace std; +; st ...

  6. PAT甲题题解-1043. Is It a Binary Search Tree (25)-二叉搜索树

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789220.html特别不喜欢那些随便转载别人的原创文章又不给 ...

  7. 【PAT甲级】1043 Is It a Binary Search Tree (25 分)(判断是否为BST的先序遍历并输出后序遍历)

    题意: 输入一个正整数N(<=1000),接下来输入N个点的序号.如果刚才输入的序列是一颗二叉搜索树或它的镜像(中心翻转180°)的先序遍历,那么输出YES并输出它的后序遍历,否则输出NO. t ...

  8. pat1043. Is It a Binary Search Tree (25)

    1043. Is It a Binary Search Tree (25) 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 ...

  9. 【PAT】1043 Is It a Binary Search Tree(25 分)

    1043 Is It a Binary Search Tree(25 分) A Binary Search Tree (BST) is recursively defined as a binary ...

随机推荐

  1. iOS数据存储之对象归档

    iOS数据存储之对象归档 对象归档 对象归档是iOS中数据持久化的一种方式. 归档是指另一种形式的序列化,但它是任何对象都可以实现的更常规的类型.使用对模型对象进行归档的技术可以轻松将复杂的对象写入文 ...

  2. MFC学习知识点20160715

    1.   sizeof()  :返回所查询目标所占用字节数 _countof() :返回所查询目标所含有元素个数 _countof 是 C++中计算一个固定大小数组长度的宏,比如: T arr[10] ...

  3. 《Genesis-3D开源游戏引擎完整实例教程-2D射击游戏篇08:弹幕系统》本系列完结

    8.弹幕系统 弹幕系统概述: 弹幕系统的设计体现了射击游戏的基本要素,玩家要在敌人放出的大量子弹(弹幕)的细小空隙间闪避,能在玩家闪躲弹幕的时候给玩家带来快感,接近满屏的子弹,增加了对玩家的视觉冲击力 ...

  4. 有关android UI 线程

    1. GUI线程框架 常见的 Swing, SWT框架都是作为单线程子系统来实现的,实际上不仅限于在Java中, Qt.MacOS Cocoa以及其他的环境中的GUI框架都是单线程的.虽然很多人尝试过 ...

  5. Codevs No.1287 矩阵乘法

    2016-06-01 16:53:23 题目链接: 矩阵乘法 (Codevs No.1287) 题目大意: 给你两个可乘矩阵a,b,求a*b 解法: 定义....... //矩阵乘法 (Codevs ...

  6. 【转】Maven实战(五)---两个war包的调用

    原博文出自于: http://blog.csdn.net/liutengteng130/article/details/42879803    感谢! 开篇前提   1.为什么要用两个war包的调用? ...

  7. 【转】Objective-C代码注释和文档输出的工具和方法

    http://blog.xcodev.com/blog/2013/11/01/code-comment-and-doc-gen-tools-for-objc/ 代码注释可以让代码更容易接受和使用,特别 ...

  8. CF#190DIV.1

    /* CF#190DIV.1-C 题意:给你n个结点的树,给这些结点标记字母AB..Z,对于标记相同的结点路径上 的结点的标记必须有一个是大于该标记的:问是否可以标记(A是最大标记) 分析:整天思路就 ...

  9. Mybatis中实体类中的字段跟对应表的字段不一致时解决办法

    解决字段名与实体类属性名不相同的冲突 实体类字段: public class Order { private int id; private String orderNo; private float ...

  10. [iOS 多线程 & 网络 - 2.5] - 小文件上传

    A.文件上传 思路: 发送文件数据给服务器 使用post请求 必须手动设置请求头: 内容大小Content-Length & 内容类型 Content-Type 请求体:文件数据 文件上传的格 ...