PAT 2019年春季 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

实现思路:
题意给出后序中序建树，并且给一些语句，判断这些说法是否正确，这是一道stl的运用题，但是也有很多的坑点，就比如在判断A是否为B的父母的时候，别用parent[A]->data==B，因为有可能A的父节点是空的情况，就会出现空指针错误。最好的解决办法应该单独用一个map容器保存每个数值所对应的node结点指针，这样就只需要判断结点指针地址是否相等即可，还有本题运用sscanf(源字符串，"所要求的格式"，&值)，将所需要的int值单独取出，更加方便。

AC代码：

#include <iostream>
#include <unordered_map>
#include <cstring>
#include <queue>
using namespace std;
const int N=50;
int post[N],in[N],high[1010];
struct node {
	int data,high;
	node *l,*r;
};
unordered_map<int,node*> parent,mp;

node* build(int postL,int postR,int inL,int inR) {
	if(postL>postR) return NULL;
	node *root=new node;
	root->data=post[postR];
	mp[root->data]=root;
	int k;
	for(k=inL; k<=inR; k++) {
		if(root->data==in[k]) break;
	}
	int leftNodeNum=k-inL;
	root->l=build(postL,postL+leftNodeNum-1,inL,k-1);
	if(root->l) parent[root->l->data]=root;
	root->r=build(postL+leftNodeNum,postR-1,k+1,inR);
	if(root->r) parent[root->r->data]=root;
	return root;
}

bool bfs(node *root) {
	queue<node*> q;
	root->high=1;
	q.push(root);
	bool flag=true;
	while(!q.empty()) {
		node *root=q.front();
		high[root->data]=root->high;
		q.pop();
		if((root->l==NULL&&root->r!=NULL)||(root->l!=NULL&&root->r==NULL)) flag=false;
		if(root->l) {
			root->l->high=root->high+1;
			q.push(root->l);
		}
		if(root->r) {
			q.push(root->r);
			root->r->high=root->high+1;
		}
	}
	return flag;
}

int main() {
	cin.tie(0);
	int n,m;
	cin>>n;
	for(int i=0; i<n; i++) scanf("%d",&post[i]);
	for(int i=0; i<n; i++) scanf("%d",&in[i]);
	node *root=build(0,n-1,0,n-1);
	cin>>m;
	bool ans=bfs(root);
	getchar();
	while(m--) {
		string str;
		getline(cin,str);
		int a,b;
		bool tag=false;
		if(str.find("root")!=string::npos) {
			sscanf(str.c_str(),"%d is the root",&a);
			if(a==root->data) tag=true;
		} else if(str.find("siblings")!=string::npos) {
			sscanf(str.c_str(),"%d and %d are siblings",&a,&b);
			if(parent[a]==parent[b]) tag=true;
		} else if(str.find("parent")!=string::npos) {
			sscanf(str.c_str(),"%d is the parent of %d",&a,&b);
			if(parent[b]==mp[a]) tag=true;
		} else if(str.find("left")!=string::npos) {
			sscanf(str.c_str(),"%d is the left child of %d",&a,&b);
			if(mp[b]->l==mp[a]) tag=true;
		} else if(str.find("right")!=string::npos) {
			sscanf(str.c_str(),"%d is the right child of %d",&a,&b);
			if(mp[b]->r==mp[a]) tag=true;
		} else if(str.find("level")!=string::npos) {
			sscanf(str.c_str(),"%d and %d are on the same level",&a,&b);
			if(high[a]==high[b]) tag=true;
		} else tag=ans;
		if(tag) printf("Yes\n");
		else printf("No\n");
	}

	return 0;
}