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:

 /*
 二叉树中各结点值为互不相同的正整数
 给出后序遍历和中序遍历

 判断所给语句是否正确
 根节点
 兄弟
 父结点
 左孩子
 右孩子
 同一层
 满二叉树

 基本思路：
 建树，存储<键值,结点>的映射，存储结点所在层次，判断是否为满二叉树
 sscanf读取字符串
 */
 #include<cstdio>
 #include<string>
 #include<unordered_map>
 #include<iostream>
 using namespace std;
 const int M=1e3+;
 struct node
 {
     int data,high;
     node *lchild,*rchild;
 };
 int in[M],pt[M],hs[M]={},fa[M]={},isfull=;
 unordered_map<int,node*> mp;

 node *Create(int ptL, int ptR, int inL, int inR, int layer, int father)
 {
     if(ptL > ptR)
         return NULL;
     node *root = new node;
     root->data = pt[ptR];
     root->high = layer;
     int k;
     for(k=inL; k<=inR; k++)
         if(in[k]==pt[ptR])
             break;
     int numLeft = k-inL;
     root->lchild = Create(ptL,ptL+numLeft-,inL,k-,layer+, root->data);
     root->rchild = Create(ptL+numLeft,ptR-,k+,inR,layer+, root->data);
     hs[root->data]=;
     fa[root->data]=father;
     mp[root->data] = root;
     if((root->lchild==NULL&&root->rchild!=NULL) || (root->lchild!=NULL&&root->rchild==NULL))
         isfull=;
     return root;
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDEG

     int n,m;
     scanf("%d", &n);
     for(int i=; i<n; i++)
         scanf("%d", &pt[i]);
     for(int i=; i<n; i++)
         scanf("%d", &in[i]);

     node *root=Create(,n-,,n-,,-);
     scanf("%d\n", &m);
     while(m--)
     {
         int v1,v2;
         string st;
         getline(cin,st);
         if(st.find("root") != string::npos)
         {
             sscanf(st.c_str(), "%d is the root", &v1);
             if(v1 == pt[n-])
                 printf("Yes\n");
             else
                 printf("No\n");
         }
         else if(st.find("siblings") != string::npos)
         {
             sscanf(st.c_str(), "%d and %d are siblings", &v1,&v2);
             if(hs[v1]== && hs[v2]== && fa[v1]==fa[v2])
                 printf("Yes\n");
             else
                 printf("No\n");
         }
         else if(st.find("parent") != string::npos)
         {
             sscanf(st.c_str(), "%d is the parent of %d", &v1,&v2);
             if(hs[v1]== && hs[v2]== && fa[v2]==v1)
                 printf("Yes\n");
             else
                 printf("No\n");
         }
         else if(st.find("left") != string::npos)
         {
             sscanf(st.c_str(), "%d is the left child of %d", &v1,&v2);
             if(hs[v1]== && hs[v2]== && mp[v2]->lchild==mp[v1])
                 printf("Yes\n");
             else
                 printf("No\n");
         }
         else if(st.find("right") != string::npos)
         {
             sscanf(st.c_str(), "%d is the right child of %d", &v1,&v2);
             if(hs[v1]== && hs[v2]== && mp[v2]->rchild==mp[v1])
                 printf("Yes\n");
             else
                 printf("No\n");
         }
         else if(st.find("same") != string::npos)
         {
             sscanf(st.c_str(), "%d and %d are on the same level", &v1,&v2);
             if(hs[v1]== && hs[v2]== && mp[v1]->high==mp[v2]->high)
                 printf("Yes\n");
             else
                 printf("No\n");
         }
         else if(st.find("full") != string::npos)
         {
             if(isfull)
                 printf("Yes\n");
             else
                 printf("No\n");
         }
     }

     return ;
 }