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 32阅读 0赞

[算法笔记总目录][Link 1]  
[关键英语单词解释][Link 2]  
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:

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 103 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

#include<bits/stdc++.h>
    using namespace std;
    struct node{ 
    	int val,level;	
    	node *parent, *lc, *rc;
    	node(int v, int lev) : val(v), parent(NULL), lc(NULL), rc(NULL), level(lev) { }
    };
    bool isFull = true;
    unordered_map<int, node*> mp;
    vector<int> post, in;
    node *root = NULL;
    node *create(int postL,int postR,int inL,int inR,int level){ 
    	if(postL > postR) return NULL;
    	node *root = new node(post[postR],level++);
    	int k;
    	for(k=inL;k<=inR;k++){ 
    		if(in[k]==post[postR])break;
    	}
    	int numleft = k-inL;
    	root->lc = create(postL,postL+numleft-1,inL,k-1,level);
    	root->rc = create(postL+numleft,postR-1,k+1,inR,level);
    	if(root->lc != NULL) root->lc->parent = root;
    	if(root->rc != NULL) root->rc->parent = root;
    	if ((root -> lc == NULL && root -> rc != NULL) || (root -> lc != NULL && root -> rc == NULL))  isFull = false;      
        mp[post[postR]] = root;
    	return root;
    }
    
    bool judge(string s) { 
        int u, v;
        if (s.find("root") != string::npos) { 
            sscanf(s.c_str(), "%d is the root", &u);
            return u == root -> val;
        } else if (s.find("siblings") != string::npos) { 
            sscanf(s.c_str(), "%d and %d are siblings", &u, &v);
            return mp[u] -> parent == mp[v] -> parent;
        } else if (s.find("parent") != string::npos) { 
            sscanf(s.c_str(), "%d is the parent of %d", &u, &v);
            return mp[v] -> parent == mp[u];
        } else if (s.find("left") != string::npos) { 
            sscanf(s.c_str(), "%d is the left child of %d", &u, &v);
            return mp[v] -> lc == mp[u];
        } else if (s.find("right") != string::npos) { 
            sscanf(s.c_str(), "%d is the right child of %d", &u, &v);
            return mp[v] -> rc == mp[u];
        } else if (s.find("level") != string::npos) { 
            sscanf(s.c_str(), "%d and %d are on the same level", &u, &v);
            return mp[u] -> level == mp[v] -> level;
        } else { 
            return isFull;
        }
    }
    int main() { 
        int n, k;
        cin >> n;
        post.resize(n);
        in.resize(n);
        for (int i = 0; i < n; i++) cin >> post[i];
        for (int i = 0; i < n; i++) cin >> in[i];
        root = create(0, n-1, 0, n-1, 1);
        cin >> k;
        getchar();
        for (int i = 0; i < k; i++) { 
            string temp;
            getline(cin, temp);
            if (judge(temp)) printf("Yes\n");
            else printf("No\n");
        }
        return 0;
    }

[Link 1]: https://blog.csdn.net/qq_21480607/article/details/106660085
[Link 2]: https://blog.csdn.net/qq_21480607/article/details/106987591