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

汇总贴

2020年3月PAT甲级满分必备刷题技巧

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），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;
}