19年春季第四题 PAT甲级 1159 Structure of a Binary Tree（30分）

柔光的暖阳◎ 2023-07-07 15:58 3阅读 0赞

# 汇总贴 #

[2020年3月PAT甲级满分必备刷题技巧][2020_3_PAT]

# 题目 #

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 (≤30), 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 10^3 and are separated by a space.

Then another positive integer M (≤30) 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: ###

9
    16 7 11 32 28 2 23 8 15
    16 23 7 32 11 2 28 15 8
    7
    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

### Sample Output: ###

Yes
    No
    Yes
    No
    Yes
    Yes
    Yes

# 题目分析 #

1.postorder and inorder traversal sequences  
给定后序和中序，这个是PAT甲级题库中很常见的知识点“二叉树的遍历”，可以参考《算法笔记》9.2节和题库的1086、1119、1138。

2.因为可能不是完全二叉树，同时输出的东西也挺复杂的，所以应该果断用DFS建二叉树。建树方法略有不同于1135（[https://www.liuchuo.net/archives/4099][https_www.liuchuo.net_archives_4099]），是采用map和结构体。

3.本题有7种输出，对应7个数据，都可以在DFS建树中体现：  
（1）root，建树的返回就是root的值  
（2）siblings，等价于两个节点的祖先是同一个  
（3）parent，每个节点存父亲的值  
（4）leftchild，A节点的左孩子是B，为了访问A的时候就能读出B节点的值，要用map和结构体  
（5）rightchild，B节点的左孩子是A  
（6）same level，记录节点的深度  
（7）full tree，记录节点有没有左右孩子

# 满分代码 #

来源致谢：https://blog.csdn.net/lianwaiyuwusheng/article/details/88084378

#include<iostream>
    #include<vector>
    #include<cstdio>
    #include<set>
    #include<map>
    #include<algorithm>
    using namespace std;
    #define MAX 40
    struct node{
        int l,r,d,p;
        node(){}
        node(int l,int r,int d,int p):l(l),r(r),d(d),p(p){}
    };
    map<int,node>arr;
    int po[MAX],in[MAX],N,A,B;
    bool flag=0;
    string str;
    int tree(int poR,int inL,int inR,int de,int pa)
    {
        if(inL>inR)return -1;
        int i=inL;
        while(in[i]!=po[poR])++i;
        arr[po[poR]].p=pa;
        arr[po[poR]].d=de;
        arr[po[poR]].l=tree(poR-(inR-i)-1,inL,i-1,de+1,po[poR]);
        arr[po[poR]].r=tree(poR-1,i+1,inR,de+1,po[poR]);
        if((arr[po[poR]].r==-1&&arr[po[poR]].l!=-1)||(arr[po[poR]].r!=-1&&arr[po[poR]].l==-1))flag=1;
        return po[poR];
    }
    int input()
    {
        int i;
        cin>>str;
        if(str[0]>='0'&&str[0]<='9'){
            i=0;A=0;
            while(i<str.length()){A=A*10+str[i]-'0';++i;}
            cin>>str;
            if(str=="and"){
                scanf("%d",&B);
                cin>>str;cin>>str;
                if(str=="siblings"){
                    return 2;
                }else{
                    i=3;
                    while(i--){cin>>str;}
                    return 6;
                }
            }else{
                cin>>str;cin>>str;
                if(str=="root")return 1;
                else if(str=="parent"){
                    cin>>str;scanf("%d",&B);
                    return 3;
                }else if(str=="left"){
                    cin>>str;cin>>str;scanf("%d",&B);
                    return 4;
                }else if(str=="right"){
                    cin>>str;cin>>str;scanf("%d",&B);
                    return 5;
                }
            }
        }else{
            i=4;
            while(i--)cin>>str;
            return 7;
        }
    }
    int main()
    {
        //freopen("test.txt","r",stdin);
        scanf("%d",&N);
        for(int i=0;i<N;++i)scanf("%d",&po[i]);
        for(int i=0;i<N;++i)scanf("%d",&in[i]);
        int root=tree(N-1,0,N-1,0,-1);
        scanf("%d",&N);
        while(N--){
            int x=input();
            bool f=0;
            if(x==1){
                if(A!=root)f=1;
            }else if(x==2){
                if(arr[A].p!=arr[B].p)f=1;
            }else if(x==3){
                if(arr[B].p!=A)f=1;
            }else if(x==4){
                if(arr[B].l!=A)f=1;
            }else if(x==5){
                if(arr[B].r!=A)f=1;
            }else if(x==6){
                if(arr[B].d!=arr[A].d)f=1;
            }else if(x==7){
                if(flag)f=1;
            }
            if(f)printf("No\n");
            else printf("Yes\n");
        }
        return 0;
    }

[2020_3_PAT]: https://blog.csdn.net/allisonshing/article/details/104213959#comments
[https_www.liuchuo.net_archives_4099]: https://www.liuchuo.net/archives/4099