PAT 1167 Cartesian Tree-优快云博客

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.

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

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

按照题目要求构建二叉树，再层次遍历即可

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int N = 31;
struct node{
	int ele=-1;
	int lchild=-1;
	int rchild=-1;
} nodes[N];
int min_heap=-1;
int n;
int gol_root;
vector<int> level_order;
void search(int fa, int l, int r)
{
	if(l>r) return;
	int root=l;
	for(int i=l+1 ; i<=r ; i++)
		if(nodes[i].ele<nodes[root].ele)
			root = i;
	if(fa == -1) gol_root = root;
	else if(nodes[fa].lchild == -1) nodes[fa].lchild=root;
	else nodes[fa].rchild=root;

	search(root, l, root-1);
	search(root, root+1, r);
}

void level_tra(int root)
{
	queue<int> q;
	q.push(root);

	while(!q.empty())
	{
		int x = q.front();
		level_order.push_back(nodes[x].ele);
		q.pop();
		if(nodes[x].lchild!=-1) q.push(nodes[x].lchild);
		if(nodes[x].rchild!=-1) q.push(nodes[x].rchild);
	}
}

int main()
{
	cin>>n;
	for(int i=1 ; i<=n ; i++)
	{
		cin>>nodes[i].ele;
	}
	search(-1,1,n);
	level_tra(gol_root);
	for(int i=0 ; i<level_order.size()-1 ; i++)
		cout<<level_order[i]<<" ";
	cout<<level_order[level_order.size()-1];
	return 0;
}