甲级PAT 2019年3月 Structure of a Binary Tree （30 分)

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 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:

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

完整代码 

#include
using namespace std;

struct Node{
	int data;
	int parent;
	int level;
	bool isleft;
}node[50];

int pre[50],in[50],post[50];
int num=0;
int n;

unordered_map premap,childnum;
void pretra(int root,int start,int end,int level,int parent,bool isleft){
	if (start>end) return;
	int i=start;
	while(i>n;
	char s[100];
	for(i=0;i>post[i];
	for(i=0;i>in[i];
	for(i=0;i<=n;i++){
		node[i].parent=-1;
		node[i].level=-1;
	}
	pretra(n-1,0,n-1,1,-1,false);
	cin>>m;
	getchar();
	bool isfulln=isfull();
	for(i=0;i15){
				if(issamelevel(l,r)) cout<<"Yes"<