PAT 1159 Structure of a Binary Tree

原题链接：PAT 1159 Structure of a Binary Tree（30分）
参考的浒鱼鱼的博客：PAT甲级 1159 Structure of a Binary Tree（30分）
关键词：dfs建树
注意：full binary tree：满二叉树 complete binary tree：完全二叉树

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（兄弟姐妹）
• 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

题目大意：给出二叉树的后序和中序序列，再给出几个陈述让你判断是否正确。
浒鱼鱼的分析： 给定后序和中序，这个是PAT甲级题库中很常见的知识点“二叉树的遍历”，可以参考《算法笔记》9.2节和题库的1086、1119、1138。

2.因为可能不是完全二叉树，同时输出的东西也挺复杂的，所以应该果断用DFS建二叉树。建树方法略有不同于1135，是采用map和结构体。

3.本题有7种输出，对应7个数据，都可以在DFS建树中体现：
（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

代码：

#include
#include
#include
#include

using namespace std;
const int maxn = 40; 
int post[maxn], in[maxn];	//后序遍历 中序遍历 

struct node{
    int l, r, depth, parent;
};

map<int, node> tree;	//编号 <-> 结点 
bool fullTree = true;	//是否是满二叉树 

int build(int postr, int inl, int inr, int dpt,int prt){	//post的右端 in的左右端 dpt深度 prt 父节点 
    if(inl > inr) return -1;
    tree[post[postr]].depth = dpt;
    tree[post[postr]].parent = prt;
    int i = inl;
    while(i < inr && in[i] != post[postr]) i++;	//在中序遍历中找到根（postr）的位置 
    
	//递归左右子树 
    tree[post[postr]].l = build(postr-inr+i-1, inl, i-1, dpt+1, post[postr]);	
    tree[post[postr]].r = build(postr-1, i+1, inr, dpt+1, post[postr]);
    
    //如果结点没有左孩子或者右孩子，就不是满二叉树 
    if(tree[post[postr]].l == -1 && tree[post[postr]].r != -1) fullTree = false;
    if(tree[post[postr]].l != -1 && tree[post[postr]].r == -1) fullTree = false;
    return post[postr];
}
int main(){
    int n, m;	//n结点数 m询问数 
    cin >> n;
    for(int i = 0; i < n; i++) scanf("%d",&post[i]);
    for(int i = 0; i<n; i++) scanf("%d",&in[i]);
    int root = build(n-1, 0, n-1, 0, -1);
    
    scanf("%d\n",&m);
    string s1, s2, s3, s4, s5, s6, s7;
    for(int i = 0; i < m; i++){
        bool flag = false;
        cin >> s1 >> s2;
        if(s2 == "and"){
            cin >> s3 >> s4 >> s5;
            if(s5=="siblings"){
                if(tree[stoi(s1)].parent == tree[stoi(s3)].parent) flag = true;
            }
			else{
                if(tree[stoi(s1)].depth == tree[stoi(s3)].depth) flag = true;
            }
        }
		else{
            cin >> s3 >> s4;
            if(s4=="root"){
                if(root == stoi(s1)) flag = true;
            }
			else if(s4 == "parent"){
                cin >> s5 >> s6;
                if(tree[stoi(s6)].parent == stoi(s1)) flag = true;
            }
			else if(s4 == "left"){
                cin >> s5 >> s6 >> s7;
                if(tree[stoi(s7)].r == stoi(s1)) flag = true;
            }
			else if(s4 == "right"){
                cin >> s5 >> s6 >> s7;
                if(tree[stoi(s7)].r == stoi(s1)) flag = true;
            }else{
                if(fullTree) flag = true;
            }
        }
        if(flag) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    return 0;
}