PAT-2019年冬季考试-甲级-7-4 Cartesian Tree (30分)超级详解不AC都难

题目：

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.

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

这道题的题眼：

1.给出的是个min-heap
最小堆，即所有父节点不大于子节点的二叉树。如果找到了目前区间的最小值，那就是个父节点。
2.给出的顺序是inorder
中序，也就是LNR。是不是想到了递归和分而治之的老方法了？
3.输出的顺序是level-order
层序。如果不建树，就是开个大数组，按照下标放；如果建树，就得用BFS遍历树了。

这道题的陷阱：

没有说是哪种二叉树，如果是最极端的那种，就是每行只有一个节点，画个草图就是：

4是3的左孩子，3是2的左孩子，2是1的左孩子。
题目说最多30个节点，如果不建树那得需要2的30次方那么大的数组了，那会遇到编译不通过。所以这道题还得建树。

我自己写的测试用例：

改变节点的数量和二叉树的内部结构。不过这道题本身是没啥坑的。

//输入1
8
60 58 38 52 8 82 25 70
//输出1
8 38 25 58 52 82 70 60

//输入2
13
8 4 2 9 5 9 10 1 11 6 12 3 7 13
//输出2
1 2 3 4 5 6 7 8 9 10 11 12 13

附上满分答案：

#include<iostream>
#include<queue>
using namespace std;
int n,k;
vector<int> v;
struct tree{
    int data;
    tree *l,*r;
}*T;
tree* check(int low,int high){
    if(low<0||low>high)return NULL;
    tree *t=new tree;
    int minn=low;
    for(int j=low;j<=high;j++){
        if(v[minn]>v[j])minn=j;
    }
    t->data=v[minn];
    t->l=check(low,minn-1);
    t->r=check(minn+1,high);
    return t;
}
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&k);
        v.push_back(k);
    }
    T=check(0,n-1);
    int flag=0;
    queue<tree*> q;
    q.push(T);
    while(!q.empty()){
        tree *t=q.front();
        if(flag==1)cout<<" ";
        else flag=1;
        cout<<t->data;
        if(t->l)q.push(t->l);
        if(t->r)q.push(t->r);
        q.pop();
    }
    return 0;
}