PAT（甲级）2019年冬季考试7-4 Cartesian Tree (30 分)

最新推荐文章于 2022-08-08 12:31:12 发布

原创最新推荐文章于 2022-08-08 12:31:12 发布 · 219 阅读

1 ·

CC 4.0 BY-SA版权

pat甲级同时被 2 个专栏收录

181 篇文章

订阅专栏

树

37 篇文章

订阅专栏

本文介绍如何根据一组不重复的数字序列构建最小堆Cartesian树，并实现该树的层级遍历。通过寻找序列中的最小值建立树结构，再利用队列完成遍历过程。

7-4 Cartesian Tree (30 分)

A Cartesian tree is a binary tree constructed from a sequence of distinct numbers. The tree is heap-ordered, and an inorder traversal returns the original sequence. For example, given the sequence { 8, 15, 3, 4, 1, 5, 12, 10, 18, 6 }, the min-heap Cartesian tree is shown by the figure.
在这里插入图片描述
Your job is to output the level-order traversal sequence of the min-heap Cartesian tree.

Input Specification:

Each input file contains one test case. Each case starts from giving a positive integer N (≤30), and then N distinct numbers in the next line, separated by a space. All the numbers are in the range of int.

Output Specification:

For each test case, print in a line the level-order traversal sequence of the min-heap Cartesian tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the beginning or the end of the line.

Sample Input:

10
8 15 3 4 1 5 12 10 18 6

Sample Output:

1 3 5 8 4 6 15 10 12 18

最小堆中每一棵子树的根节点都是整棵子树的最小节点，根据中序序列找到遍历序列中最小节点的索引然后建树，然后用queue遍历就行

#include<bits/stdc++.h>
using namespace std;
struct Node {
	int l = -1, r = -1;
};
vector<int> in;
vector<Node> Tree(40);
int bfs(int inL, int inR) {
	if (inL > inR) return -1;
	int Min = in[inL], V = inL;
	for (int i = inL + 1; i <= inR; i++) {
		if (in[i] < Min) {
			Min = in[i];
			V = i;
		}
	}
	Tree[V].l = bfs(inL, V - 1);
	Tree[V].r = bfs(V + 1, inR);
	return V;
}
int main() {
	int N;
	scanf("%d", &N);
	in.resize(N + 1); 
	Tree.resize(N + 1);
	for (int i = 1; i<= N; i++){
		scanf("%d", &in[i]);
	}
	int root = bfs(1, N), flag = 0;
	queue<int> Q;
	Q.push(root);
	while(!Q.empty()) {
		int temp = Q.front();
		Q.pop();
		if (flag == 0) {
			flag = 1;
		} else printf(" ");
		printf("%d", in[temp]);
		if (Tree[temp].l != -1) Q.push(Tree[temp].l);
		if (Tree[temp].r != -1) Q.push(Tree[temp].r);		
	}
	return 0;
}