【PTA-A】1043 Is It a Binary Search Tree (25 分)（前序/后续遍历、镜像遍历、BST）

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本文探讨了如何判断一个整数序列是否为二叉搜索树(BST)或其镜像的预遍历序列。通过构建BST并比较序列，文章提供了确定序列性质的方法，并展示了如何生成相应的后遍历序列。

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

Sample Output 1:

YES
5 7 6 8 11 10 8

Sample Input 2:

7
8 10 11 8 6 7 5

Sample Output 2:

YES
11 8 10 7 5 6 8

Sample Input 3:

7
8 6 8 5 10 9 11

Sample Output 3:

NO

#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
struct Node {
	int data;
	Node *left, *right;
};
vector<int> origin, pre, preM, post, postM;

void insert(Node* &root, int data) {
	if (root == NULL) {
		root = new Node;
		root->data = data;
		root->left = root->right = NULL;
		return;
	}
	if (data < root->data)insert(root->left, data);
	else insert(root->right,data);
}

void preorder(Node* root, vector<int>& vi) {
	if (root == NULL)return;
	vi.push_back(root->data);
	preorder(root->left, vi);
	preorder(root->right, vi);
}
void preorderM(Node* root, vector<int>& vi) {
	if (root == NULL)return;
	vi.push_back(root->data);
	preorderM(root->right, vi);
	preorderM(root->left, vi);
}
void postorder(Node* root, vector<int>& vi) {
	if (root == NULL)return;
	postorder(root->left, vi);
	postorder(root->right, vi);
	vi.push_back(root->data);
}
void postorderM(Node* root, vector<int>& vi) {
	if (root == NULL)return;
	postorderM(root->right, vi);
	postorderM(root->left, vi);
	vi.push_back(root->data);
}

int main() {
	int n, data;
	cin >> n;
	Node* root = NULL;
	for (int i = 0; i < n; i++) {
		scanf("%d", &data);
		origin.push_back(data);
		insert(root, data);
	}
	preorder(root, pre);
	preorderM(root, preM);
	postorder(root, post);
	postorderM(root, postM);
	if (origin == pre) {
		printf("YES\n");
		for (int i = 0; i < post.size(); i++) {
			printf("%d", post[i]);
			if (i != post.size() - 1)printf(" ");
		}
	}
	else if (origin == preM) {
		printf("YES\n");
		for (int i = 0; i < postM.size(); i++) {
			printf("%d", postM[i]);
			if (i != postM.size() - 1)printf(" ");
		}
	}
	else { printf("NO\n"); }
	return 0;
}