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 /*
  2 二叉树中各结点值为互不相同的正整数
  3 给出后序遍历和中序遍历
  4 
  5 判断所给语句是否正确
  6 根节点
  7 兄弟
  8 父结点
  9 左孩子
 10 右孩子
 11 同一层
 12 满二叉树
 13 
 14 基本思路：
 15 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
 16 sscanf读取字符串
 17 */
 18 #include<cstdio>
 19 #include<string>
 20 #include<unordered_map>
 21 #include<iostream>
 22 using namespace std;
 23 const int M=1e3+10;
 24 struct node
 25 {
 26     int data,high;
 27     node *lchild,*rchild;
 28 };
 29 int in[M],pt[M],hs[M]={0},fa[M]={0},isfull=1;
 30 unordered_map<int,node*> mp;
 31 
 32 node *Create(int ptL, int ptR, int inL, int inR, int layer, int father)
 33 {
 34     if(ptL > ptR)
 35         return NULL;
 36     node *root = new node;
 37     root->data = pt[ptR];
 38     root->high = layer;
 39     int k;
 40     for(k=inL; k<=inR; k++)
 41         if(in[k]==pt[ptR])
 42             break;
 43     int numLeft = k-inL;
 44     root->lchild = Create(ptL,ptL+numLeft-1,inL,k-1,layer+1, root->data);
 45     root->rchild = Create(ptL+numLeft,ptR-1,k+1,inR,layer+1, root->data);
 46     hs[root->data]=1;
 47     fa[root->data]=father;
 48     mp[root->data] = root;
 49     if((root->lchild==NULL&&root->rchild!=NULL) || (root->lchild!=NULL&&root->rchild==NULL))
 50         isfull=0;
 51     return root;
 52 }
 53 
 54 int main()
 55 {
 56 #ifdef ONLINE_JUDGE
 57 #else
 58     freopen("Test.txt", "r", stdin);
 59 #endif // ONLINE_JUDEG
 60 
 61     int n,m;
 62     scanf("%d", &n);
 63     for(int i=0; i<n; i++)
 64         scanf("%d", &pt[i]);
 65     for(int i=0; i<n; i++)
 66         scanf("%d", &in[i]);
 67 
 68     node *root=Create(0,n-1,0,n-1,1,-1);
 69     scanf("%d\n", &m);
 70     while(m--)
 71     {
 72         int v1,v2;
 73         string st;
 74         getline(cin,st);
 75         if(st.find("root") != string::npos)
 76         {
 77             sscanf(st.c_str(), "%d is the root", &v1);
 78             if(v1 == pt[n-1])
 79                 printf("Yes\n");
 80             else
 81                 printf("No\n");
 82         }
 83         else if(st.find("siblings") != string::npos)
 84         {
 85             sscanf(st.c_str(), "%d and %d are siblings", &v1,&v2);
 86             if(hs[v1]==1 && hs[v2]==1 && fa[v1]==fa[v2])
 87                 printf("Yes\n");
 88             else
 89                 printf("No\n");
 90         }
 91         else if(st.find("parent") != string::npos)
 92         {
 93             sscanf(st.c_str(), "%d is the parent of %d", &v1,&v2);
 94             if(hs[v1]==1 && hs[v2]==1 && fa[v2]==v1)
 95                 printf("Yes\n");
 96             else
 97                 printf("No\n");
 98         }
 99         else if(st.find("left") != string::npos)
100         {
101             sscanf(st.c_str(), "%d is the left child of %d", &v1,&v2);
102             if(hs[v1]==1 && hs[v2]==1 && mp[v2]->lchild==mp[v1])
103                 printf("Yes\n");
104             else
105                 printf("No\n");
106         }
107         else if(st.find("right") != string::npos)
108         {
109             sscanf(st.c_str(), "%d is the right child of %d", &v1,&v2);
110             if(hs[v1]==1 && hs[v2]==1 && mp[v2]->rchild==mp[v1])
111                 printf("Yes\n");
112             else
113                 printf("No\n");
114         }
115         else if(st.find("same") != string::npos)
116         {
117             sscanf(st.c_str(), "%d and %d are on the same level", &v1,&v2);
118             if(hs[v1]==1 && hs[v2]==1 && mp[v1]->high==mp[v2]->high)
119                 printf("Yes\n");
120             else
121                 printf("No\n");
122         }
123         else if(st.find("full") != string::npos)
124         {
125             if(isfull)
126                 printf("Yes\n");
127             else
128                 printf("No\n");
129         }
130     }
131 
132     return 0;
133 }

 

 

 

 

 
posted @ 2019-09-01 19:41  林東雨  阅读(416)  评论(0)  编辑  收藏  举报