2019PAT春季考试第4题 7-4 Structure of a Binary Tree （30 分）

题外话：考试的时候花了一个小时做了27分，由于Siblings这个单词不知道意思，所以剩下的3分就没去纠结了，后来发现单词是兄弟的意思，气哭~~

这道题的麻烦之处在于如何从一个字符串中去找数字。先首先我们需要整行读取（包括空格），这个我自己写另一个函数，挺好用的：

string mygets()
{
    char str[];
    fgets(str,sizeof(str),stdin);
    int len = strlen(str);
    if(str[len-]=='\n') str[len-] = '\0';
    return str;
}

用的时候只要 string str = mygets();就可以了，里面已经去掉换行符了。

下一步是再写一个从istart位置读取数字的函数，先从istart位置读取字符串（遇到空格，读取结束），然后用atoi把字符串转换成数字。

//字符串中istart位置开始取某个单词（以空格结束）
int getNum(string str,int istart)
{
    int len = str.length();
    string strA="";
    for(int i=istart;i<len;++i)
        if(str[i]==' ') break;
        else strA += str[i];
    //printf("%s\n",strA.c_str());
    int a = atoi(strA.c_str());
    return a;
}

题目中的字符串格式太多了是吧，眼花缭乱：
15 is the root
8 and 2 are siblings
32 is the parent of 11
23 is the left child of 16
28 is the right child of 2
7 and 11 are on the same level
It is a full tree
拿字符串siblings举例吧，想用find函数siblings字符串，如果返回不等于-1，表示就是此种类型，然后用以下代码获取数字8和2。注意find函数的用法，返回的是查找字符串的首地址，所以find "of"返回的是o出现的首地址，没有找到of则返回-1，否则返回o的首地址。然后我们可以想象一下istart+1是f，istart+2是空格，istart+3就是我们所要的数字出现的首地址。   

    if(str.find("siblings")!=-)
    {
        int a = getNum(str,);
        int b = getNum(str,str.find("of")+);
    }

　　以下是我的代码：

PS：如果我的构树代码看不懂，可以看我写的这篇文章让你懂得已知先序（后序）与中序遍历如何快速构树：https://www.cnblogs.com/jlyg/p/10402919.html

树的结构体中我加了level变量，记录当前节点的水平深度。其他看代码吧~~

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
#include<cstring>
using namespace std;
struct Node{
    int val;
    int level;
    Node* left;
    Node* right;
    Node(){
        left = right = NULL;
        level = ;
    }
};
map<int,int> valtoid;               //通过值去找id
map<int,Node*> valtonode;           //通过值去找节点
//构造数
Node* Insert(Node* node,int val)
{
    if(node==NULL)
    {
        node = new Node();
        node->val= val;
        valtonode[val] = node;
    }
    else if(valtoid[val] < valtoid[node->val])
    {
               node->left = Insert(node->left,val);
            node->left->level = node->level+;
    }
    else
    {
        node->right= Insert(node->right,val);
        node->right->level = node->level+;
    }
    return node;
}
Node* root = NULL;
int n;
//是不是满二叉树
bool isFullTree(Node* root)
{
    queue<Node*> q;
    q.push(root);
    Node* node;
    while(!q.empty())
    {
        node = q.front();
        q.pop();
        if(node->left)
            q.push(node->left);
        if(node->right)
            q.push(node->right);
        if(node->left&&!node->right || !node->left&&node->right)
            return false;
    }
    return true;
}
//字符串中istart位置开始取某个单词（以空格结束）
int getNum(string str,int istart)
{
    int len = str.length();
    string strA="";
    for(int i=istart;i<len;++i)
        if(str[i]==' ') break;
        else strA += str[i];
    //printf("%s\n",strA.c_str());
    int a = atoi(strA.c_str());
    return a;
}
//是不是兄弟节点
bool isSiblings(Node* node,Node* child1,Node* child2)
{
    if(node==NULL) return false;
    if(node->left==child1&&node->right!=child2||
       node->left==child2&&node->right!=child1) return false;
    else if(node->left==child1&&node->right==child2||
       node->left==child2&&node->right==child1) return true;
    return isSiblings(node->left,child1,child2)||isSiblings(node->right,child1,child2);
}
//判断这一个这句话是不是对的
bool isRight(string str)
{

    if(str.find("root")!=-)
    {
        int a = getNum(str,);
        return a == root->val;
    }
    else if(str.find("siblings")!=-)
    {
        int a = getNum(str,);
        int b = getNum(str,str.find("of")+);
        //兄弟节点，必须在同一层
        if(valtonode[b]&&valtonode[a]&&valtonode[b]->level==valtonode[a]->level)
        {
            if(isSiblings(root,valtonode[a],valtonode[b]))
                return true;
        }
        return false;
    }
    else if(str.find("parent")!=-)
    {
        int a = getNum(str,);
        int b = getNum(str,str.find("of")+);
        if(valtonode[b]&&valtonode[a])
        {
            if(valtonode[a]->left == valtonode[b] || valtonode[a]->right == valtonode[b])
            return true;
        }
        return false;
    }
    else if(str.find("left")!=-)
    {
        int a = getNum(str,);
        int b = getNum(str,str.find("of")+);
        if(valtonode[b]&&valtonode[a])
        {
            if(valtonode[b]->left==valtonode[a])
                return true;
        }
        return false;
    }
    else if(str.find("right")!=-)
    {
        int a = getNum(str,);
        int b = getNum(str,str.find("of")+);
        if(valtonode[b]&&valtonode[a])
        {
            if(valtonode[b]->right==valtonode[a])
                return true;
        }
    }
    else if(str.find("same")!=-)
    {
        int a = getNum(str,);
        int b = getNum(str,str.find("and")+);
        if(valtonode[b]&&valtonode[a])
        {
            if(valtonode[b]->level==valtonode[a]->level)
                return true;
        }
        return false;
    }
    else if(str.find("full")!=-)
    {
        return isFullTree(root);
    }
    return false;
}
string mygets()
{
    char str[];
    fgets(str,sizeof(str),stdin);
    int len = strlen(str);
    if(str[len-]=='\n') str[len-] = '\0';
    return str;
}
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
    #endif
    scanf("%d",&n);
    int post[n];
    for(int i=;i<n;++i)
        scanf("%d",&post[i]);
    for(int i=;i<n;++i)
    {
        int a;
        scanf("%d",&a);
        valtoid[a] = i;
    }
    for(int i=n-;i>=;--i)
        root = Insert(root,post[i]);
    int m;
    scanf("%d",&m);
    getchar();
    for(int i=;i<m;++i)
    {
        string str = mygets();
        bool bRight = isRight(str);
        printf("%s\n",bRight?"Yes":"No");
    }
    return ;
}

题目在下面：

- Structure of a Binary Tree （ 分）
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, a binary tree can be uniquely determined.

Now given a sequence of statements about the structure of the resulting tree, you are supposed to tell if they are correct or not. A statment is one of the following:

A is the root
A and B are siblings
A is the parent of B
A is the left child of B
A is the right child of B
A and B are on the same level
It is a full tree
Note:

Two nodes are on the same level, means that they have the same depth.
A full binary tree is a tree in which every node other than the leaves has two children.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are no more than
​
​​  and are separated by a space.

Then another positive integer M (≤) is given, followed by M lines of statements. It is guaranteed that both A and B in the statements are in the tree.

Output Specification:
For each statement, print in a line Yes if it is correct, or No if not.

Sample Input:

 is the root
 and  are siblings
 is the parent of
 is the left child of
 is the right child of
 and  are on the same level
It is a full tree

Sample Output:
Yes
No
Yes
No
Yes
Yes
Yes