PAT 2019-3 7-4 Structure of a Binary Tree

Description:

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 1 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:

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

Keys:

• 二叉树的存储和遍历

Attention:

• sscanf(s.c_str(), "%d %*s %d", &a, &b);
• s.find("s") != string::npos;
• map<int,node*> mp;

Code:

 1 #include<cstdio>
 2 #include<string>
 3 #include<map>
 4 #include<iostream>
 5 using namespace std;
 6 const int M=1e4;
 7 struct node
 8 {
 9     int data;
10     node *lchild,*rchild;
11 };
12 int n,pt[M],in[M],layer[M],isfull=1,fa[M];
13 map<int,node*> mp;
14 
15 node *Create(int inL, int inR, int ptL, int ptR, int high, int father)
16 {
17     if(inL > inR)
18         return NULL;
19     node *root = new node;
20     root->data = pt[ptR];
21     fa[root->data]=father;
22     layer[root->data]=high;
23     int k;
24     for(k=inL; k<=inR; k++)
25         if(in[k] == pt[ptR])
26             break;
27     int numLeft = k-inL;
28     root->lchild = Create(inL,k-1,ptL,ptL+numLeft-1,high+1,root->data);
29     root->rchild = Create(k+1,inR,ptL+numLeft,ptR-1,high+1,root->data);
30     if((root->lchild==NULL&&root->rchild!=NULL)||(root->lchild!=NULL&&root->rchild==NULL))
31         isfull=0;
32     mp[root->data]=root;
33     return root;
34 }
35 
36 int main()
37 {
38 #ifdef ONLINE_JUDGE
39 #else
40     freopen("Test.txt", "r", stdin);
41 #endif // ONLINE_JUDGE
42 
43     int n,m,v1,v2;
44     scanf("%d", &n);
45     for(int i=0; i<n; i++)
46         scanf("%d", &pt[i]);
47     for(int i=0; i<n; i++)
48         scanf("%d", &in[i]);
49     node *root = new node;
50     root = Create(0,n-1,0,n-1,1,-1);
51     scanf("%d\n", &m);
52     string s;
53     for(int i=0; i<m; i++)
54     {
55         getline(cin,s);
56         if(s.find("root")!=string::npos)
57         {
58             sscanf(s.c_str(),"%d is the root", &v1);
59             if(v1 == pt[n-1])   printf("Yes\n");
60             else    printf("No/n");
61         }
62         else if(s.find("siblings")!=string::npos)
63         {
64             sscanf(s.c_str(),"%d and %d are siblings", &v1,&v2);
65             if(fa[v1]==fa[v2])  printf("Yes\n");
66             else    printf("No\n");
67         }
68         else if(s.find("parent")!=string::npos)
69         {
70             sscanf(s.c_str(),"%d is the parent of %d", &v1,&v2);
71             if(fa[v2]==v1)  printf("Yes\n");
72             else    printf("No\n");
73         }
74         else if(s.find("left")!=string::npos)
75         {
76             sscanf(s.c_str(),"%d is the left child of %d", &v1,&v2);
77             if(mp[v2]->lchild!=NULL && mp[v2]->lchild->data==v1) printf("Yes\n");
78             else    printf("No\n");
79         }
80         else if(s.find("right")!=string::npos)
81         {
82             sscanf(s.c_str(),"%d is the right child of %d", &v1,&v2);
83             if(mp[v2]->rchild!=NULL && mp[v2]->rchild->data==v1) printf("Yes\n");
84             else    printf("No\n");
85         }
86         else if(s.find("same")!=string::npos)
87         {
88             sscanf(s.c_str(),"%d and %d are on the same level", &v1,&v2);
89             if(layer[v1]==layer[v2])    printf("Yes\n");
90             else    printf("No\n");
91         }
92         else if(s.find("full")!=string::npos)
93             if(isfull)  printf("Yes\n");
94             else    printf("No\n");
95     }
96 
97     return 0;
98 }