PAT 甲级 刷题日记|A 1043 Is It a Binary Search Tree (25 分)

单词积累

• Binary Search Tree 二叉搜索树
• recursively 递归地

题目

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line YES if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or NO if not. Then if the answer is YES, print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

7
8 6 5 7 10 8 11
结尾无空行

Sample Output 1:

YES
5 7 6 8 11 10 8
结尾无空行

Sample Input 2:

7
8 10 11 8 6 7 5
结尾无空行

Sample Output 2:

YES
11 8 10 7 5 6 8
结尾无空行

Sample Input 3:

7
8 6 8 5 10 9 11
结尾无空行

Sample Output 3:

NO
结尾无空行

思路

涉及到的知识点是二叉排序树的建立和遍历。镜像树并不需要再次建立，改变访问顺序即可。比较巧妙的是晴神将遍历顺序用vector存储，可以方便地用 == 比较，而不需要逐个遍历元素比较，学到了。（但据说这种用法已经没c++20删除）。

考察基本模板的题目，现在做的还是太慢了。

代码

#include 
using namespace std;

int num;
vector ori;
vector pre;
vector post;
vector mipre;
vector mipost;


struct node{
    int data;
    node* leftchild;
    node* rightchild;
    node(int d): data(d), leftchild(NULL), rightchild(NULL) {
    }
};

node* insert(node* root, int x) {
    if(root == NULL) {
        root = new node(x);
    } else if (x < root->data) {
        root->leftchild = insert(root->leftchild, x);
    } else {
        root->rightchild = insert(root->rightchild, x);
    }
    return root;
}

void preOrder(node* root) {
    if (root == NULL) return;
    pre.push_back(root->data);
    preOrder(root->leftchild);
    preOrder(root->rightchild);
}

void postOrder(node* root) {
    if (root == NULL) return;
    postOrder(root->leftchild);
    postOrder(root->rightchild);
    post.push_back(root->data);
}

void MirrorPreOrder(node* root) {
    if (root == NULL) return;
    mipre.push_back(root->data);
    MirrorPreOrder(root->rightchild);
    MirrorPreOrder(root->leftchild);
}

void MirrorPostOrder(node* root) {
    if (root == NULL) return;
    MirrorPostOrder(root->rightchild);
    MirrorPostOrder(root->leftchild);
    mipost.push_back(root->data);
}

int main() {
    cin>>num;
    node* root = NULL; 
    int d;
    for (int i = 0; i < num; i++) {
        cin>>d;
        ori.push_back(d);
        root = insert(root, d);
    }
    preOrder(root);
    MirrorPreOrder(root);
    
    if (ori == pre) {
        cout<<"YES"<