PAT甲级2019春季7-4 1159.Structure of a Binary Tree (30 分)

￡神魔★判官ぃ 2023-02-23 12:09 146阅读 0赞

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关键英语单词解释
7-4 Structure of a Binary Tree (30 分)

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:

  1. A is the root
  2. A and B are siblings
  3. A is the parent of B
  4. A is the left child of B
  5. A is the right child of B
  6. A and B are on the same level
  7. 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:

  1. 9
  2. 16 7 11 32 28 2 23 8 15
  3. 16 23 7 32 11 2 28 15 8
  4. 7
  5. 15 is the root
  6. 8 and 2 are siblings
  7. 32 is the parent of 11
  8. 23 is the left child of 16
  9. 28 is the right child of 2
  10. 7 and 11 are on the same level
  11. It is a full tree

Sample Output:

  1. Yes
  2. No
  3. Yes
  4. No
  5. Yes
  6. Yes
  7. Yes

题意

后序和中序建树，并判断
（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

注：代码仅通过测试样例

  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. struct node{ 
  4.     int val,level;    
  5.     node *parent, *lc, *rc;
  6.     node(int v, int lev) : val(v), parent(NULL), lc(NULL), rc(NULL), level(lev) { }
  7. };
  8. bool isFull = true;
  9. unordered_map<int, node*> mp;
  10. vector<int> post, in;
  11. node *root = NULL;
  12. node *create(int postL,int postR,int inL,int inR,int level){ 
  13.     if(postL > postR) return NULL;
  14.     node *root = new node(post[postR],level++);
  15.     int k;
  16.     for(k=inL;k<=inR;k++){ 
  17.         if(in[k]==post[postR])break;
  18.     }
  19.     int numleft = k-inL;
  20.     root->lc = create(postL,postL+numleft-1,inL,k-1,level);
  21.     root->rc = create(postL+numleft,postR-1,k+1,inR,level);
  22.     if(root->lc != NULL) root->lc->parent = root;
  23.     if(root->rc != NULL) root->rc->parent = root;
  24.     if ((root -> lc == NULL && root -> rc != NULL) || (root -> lc != NULL && root -> rc == NULL))  isFull = false;      
  25.     mp[post[postR]] = root;
  26.     return root;
  27. }
  29. bool judge(string s) { 
  30.     int u, v;
  31.     if (s.find("root") != string::npos) { 
  32.         sscanf(s.c_str(), "%d is the root", &u);
  33.         return u == root -> val;
  34.     } else if (s.find("siblings") != string::npos) { 
  35.         sscanf(s.c_str(), "%d and %d are siblings", &u, &v);
  36.         return mp[u] -> parent == mp[v] -> parent;
  37.     } else if (s.find("parent") != string::npos) { 
  38.         sscanf(s.c_str(), "%d is the parent of %d", &u, &v);
  39.         return mp[v] -> parent == mp[u];
  40.     } else if (s.find("left") != string::npos) { 
  41.         sscanf(s.c_str(), "%d is the left child of %d", &u, &v);
  42.         return mp[v] -> lc == mp[u];
  43.     } else if (s.find("right") != string::npos) { 
  44.         sscanf(s.c_str(), "%d is the right child of %d", &u, &v);
  45.         return mp[v] -> rc == mp[u];
  46.     } else if (s.find("level") != string::npos) { 
  47.         sscanf(s.c_str(), "%d and %d are on the same level", &u, &v);
  48.         return mp[u] -> level == mp[v] -> level;
  49.     } else { 
  50.         return isFull;
  51.     }
  52. }
  53. int main() { 
  54.     int n, k;
  55.     cin >> n;
  56.     post.resize(n);
  57.     in.resize(n);
  58.     for (int i = 0; i < n; i++) cin >> post[i];
  59.     for (int i = 0; i < n; i++) cin >> in[i];
  60.     root = create(0, n-1, 0, n-1, 1);
  61.     cin >> k;
  62.     getchar();
  63.     for (int i = 0; i < k; i++) { 
  64.         string temp;
  65.         getline(cin, temp);
  66.         if (judge(temp)) printf("Yes\n");
  67.         else printf("No\n");
  68.     }
  69.     return 0;
  70. }

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