1159.Structure of a Binary Tree (30 分)（二叉树，map，中序后序建树-顺便求高，sscanf的使用）

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

/*
 * @Descripttion: 
 * @version: 
 * @Author: iDestro
 * @Date: 2019-09-01 08:26:35
 * @LastEditors: iDestro
 * @LastEditTime: 2019-09-01 14:05:27
 */
#include <iostream>
#include <vector>
#include <unordered_map>
#include <cctype>
#include <queue>
using namespace std;
vector<int> post, in;
struct Node {
    int val, height;
    Node *parent, *left, *right;
    Node(int x, int h) {
        this -> val = x;
        this -> height = h;
        this -> left = this -> right = this -> parent = NULL;
    }
};

unordered_map<int, Node*> mp;

bool isfull = true;

// 中序后序建树
Node *create(int postL, int postH, int inL, int inH, int height) {
    if (postL > postH) {
        return NULL;
    }
    // 分配节点并赋值高度
    Node *root = new Node(post[postH], height);
    int k;
    for (k = inL; in[k] != post[postH]; k++);
    int leftNum = k - inL;
    root -> left = create(postL, postL+leftNum-1, inL, inL+leftNum-1, height+1);
    root -> right = create(postL+leftNum, postH-1, inL+leftNum+1, inH, height+1);
    // 设置孩子节点的父亲指针
    if (root -> left != NULL) root -> left -> parent = root;
    if (root -> right != NULL) root -> right -> parent = root;
    // 判断每个非叶节点是否有两个孩子
    if ((root -> left == NULL && root -> right != NULL) || (root -> left != NULL && root -> right == NULL)) {
        isfull = false;
    }
    // 存储<关键值,节点索引>
    mp[post[postH]] = root;
    return root;
}

Node *root = NULL;
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] -> left == 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] -> right == 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] -> height == mp[v] -> height;
    } 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, 0);
    cin >> k;
    getchar();
    for (int i = 0; i < k; i++) {
        string temp;
        getline(cin, temp);
        if (judge(temp)) {
            cout << "Yes" << endl;
        } else {
            cout << "No" << endl;
        }
    }
    return 0;
}