Given a tree, you are supposed to tell if it is a complete binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N−1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a - will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each case, print in one line YES and the index of the last node if the tree is a complete binary tree, or NO and the index of the root if not. There must be exactly one space separating the word and the number.

Sample Input 1:

9
7 8
- -
- -
- -
0 1
2 3
4 5
- -
- -

Sample Output 1:

YES 8

Sample Input 2:

8
- -
4 5
0 6
- -
2 3
- 7
- -
- -

Sample Output 2:

NO 1
/*
思路是先建树,找根节点,先序遍历找到最大节点,如果等于n表示完全二叉树
*/
#include<iostream>
#include<cstdlib>
using namespace std;
const int maxn = ; bool isRoot[maxn] = {};
int max_index = -,last_v;
int root = ; struct Node
{
int left,right;
}node[maxn]; int getNum(string s)
{
int iRet = -;
if(s != "-")
{
iRet = atoi(s.c_str());
isRoot[iRet] = ;
}
return iRet;
} void FindRoot(int n)
{
for(int i = ; i < n; i++)
{
if( == isRoot[i])
{
root = i;
break;
}
}
} void preOrder(int v,int index)
{
if(- == v)
{
return;
}
if(index > max_index)
{
max_index = index;
last_v = v;
}
preOrder(node[v].left,index*);
preOrder(node[v].right,index*+);
} int main()
{
int n;
cin >> n;
string s1,s2;
for(int i = ; i < n; i++)
{
cin >> s1 >> s2;
node[i].left = getNum(s1);
node[i].right = getNum(s2);
}
FindRoot(n);
preOrder(root,);
if(max_index == n)
{
printf("YES %d",last_v);
}
else
{
printf("NO %d",root);
}
return ;
}

下面版本三个点段错误,待查

#include<cstdio>

const int maxn = ;
struct Node
{
int left,right;
}node[maxn]; int max_index = -,last_v;
bool isRoot[maxn] = {};
int root; int changeNum(char c)
{
int iRet = -;
if(c != '-')
{
iRet = c - '';
isRoot[iRet] = ;
}
return iRet;
} void FindRoot(int n)
{
for(int i = ; i < n; i++)
{
if( == isRoot[i])
{
root = i;
break;
}
}
} void preOrder(int v,int index)
{
if(- == v)
{
return;
}
if(index > max_index)
{
max_index = index;
last_v = v;
}
preOrder(node[v].left,index*);
preOrder(node[v].right,index*+);
} int main()
{
int n;
scanf("%d",&n);
char c1,c2;
for(int i = ; i < n; i++)
{
getchar();
scanf("%c%*c%c",&c1,&c2);
node[i].left = changeNum(c1);
node[i].right = changeNum(c2);
}
FindRoot(n);
preOrder(root,);
if(max_index == n)
{
printf("YES %d",last_v);
}
else
{
printf("NO %d",root);
}
return ;
}

1110 Complete Binary Tree (25 分)的更多相关文章

  1. 【PAT甲级】1110 Complete Binary Tree (25分)

    题意: 输入一个正整数N(<=20),代表结点个数(0~N-1),接着输入N行每行包括每个结点的左右子结点,'-'表示无该子结点,输出是否是一颗完全二叉树,是的话输出最后一个子结点否则输出根节点 ...

  2. [二叉树建树&完全二叉树判断] 1110. Complete Binary Tree (25)

    1110. Complete Binary Tree (25) Given a tree, you are supposed to tell if it is a complete binary tr ...

  3. 1110. Complete Binary Tree (25)

    Given a tree, you are supposed to tell if it is a complete binary tree. Input Specification: Each in ...

  4. PAT Advanced 1110 Complete Binary Tree (25) [完全⼆叉树]

    题目 Given a tree, you are supposed to tell if it is a complete binary tree. Input Specification: Each ...

  5. PAT甲题题解-1110. Complete Binary Tree (25)-(判断是否为完全二叉树)

    题意:判断一个节点为n的二叉树是否为完全二叉树.Yes输出完全二叉树的最后一个节点,No输出根节点. 建树,然后分别将该树与节点树为n的二叉树相比较,统计对应的节点个数,如果为n,则为完全二叉树,否则 ...

  6. PAT (Advanced Level) 1110. Complete Binary Tree (25)

    判断一棵二叉树是否完全二叉树. #include<cstdio> #include<cstring> #include<cmath> #include<vec ...

  7. PAT甲级——1110 Complete Binary Tree (完全二叉树)

    此文章同步发布在CSDN上:https://blog.csdn.net/weixin_44385565/article/details/90317830   1110 Complete Binary ...

  8. 1110 Complete Binary Tree

    1110 Complete Binary Tree (25)(25 分) Given a tree, you are supposed to tell if it is a complete bina ...

  9. 1110 Complete Binary Tree (25 分)

    1110 Complete Binary Tree (25 分) Given a tree, you are supposed to tell if it is a complete binary t ...

随机推荐

  1. Idea导入Eclipse的Web项目并部署到Tomcat

    ⒈启动Idea,选择导入项目 选择导入的项目路径后,选择项目类型后一路next即可. ⒉选择File->Project Structure打开项目配置窗口(ctrl + alt + shift ...

  2. 关于maven自动部署tomcat9 步骤

    maven 自动部署tomcat9 (远程方法) 1.首先要去配置用户,在tomcat的conf中有tomcat_users.xml,在其中有tomcat-user的配置 配置:<tommcat ...

  3. 用Java实现对英文版《飘》的文件读取与写入操作

    从文件读入<飘>的英文版,并将结果输出到文件中 要求一: 实现对英文版<飘>的字母出现次数统计 package File; import java.io.FileInputSt ...

  4. 摘抄大神对VUE 中slot-scope的深度理解

    Vue的slot-scope的场景的个人理解 这篇文章不是单纯把文档的话和api拿来翻译和演示,而是谈谈我对于slot-scope的使用场景的个人理解,如果理解错误,欢迎讨论! Vue的插槽slot, ...

  5. Eclipse 新建.jsp页面后,页面头部标签报错的解决方法

    Eclipse 新建.jsp页面后,页面头部标签报错的解决方法 1.报错地方: 2.解决方法: .jsp页面右键==>BUild Path ==>Configure Build Path. ...

  6. Github 添加公匙 出错 (我真傻 真的)

    网上一搜一箩筐 之前配了很多次都没问题 重装系统后配了半天总是提示 github Key is invalid. You must supply a key in OpenSSH public key ...

  7. Luogu P4436 [HNOI/AHOI2018]游戏

    题目 我们要求出\(l_i,r_i\)表示\(i\)最远能够到达的最左边和最右边的格子. 首先有一个比较简单的暴力,就是每次我们选择一个格子,然后从当前格子开始往左右暴力扩展,找到能够到达的最远的格子 ...

  8. 谷歌官方颜色库 MaterialDesignColor

    谷歌官方颜色库 MaterialDesignColor

  9. 使用Python基于VGG/CTPN/CRNN的自然场景文字方向检测/区域检测/不定长OCR识别

    GitHub:https://github.com/pengcao/chinese_ocr https://github.com/xiaofengShi/CHINESE-OCR |-angle 基于V ...

  10. tasks.json 配置 解决vscode控制台乱码问题

    { "version": "2.0.0", "command": "dotnet", "tasks" ...