PAT-2019年冬季考试-甲级 7-4 Cartesian Tree (30分)（最小堆的中序遍历求层序遍历，递归建树bfs层序）

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 (≤), 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

题意：

给一个最小堆的中序遍历，求层序遍历。

题解：

也不难，最朴素稳妥的方法就是老老实实建树，老老实实bfs层序遍历。最小堆的叶节点总是小于等于根节点，所以每次都挑出最小的值最为根，然后左子树右子树重复上述过程即可。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int a[],b[];
int n;
struct node{
    int d;
    node *left,*right;
};
queue<node*>q;
vector<int>v;
node* build(int l,int r){
    if(l>r) return NULL;
    node *root=new node();
    int pos=-,k=;//找到最小的
    for(int i=l;i<=r;i++){
        if(k>a[i]){
            k=a[i];
            pos=i;
        }
    }
    root->d=a[pos];
    root->left=build(l,pos-);//左子树
    root->right=build(pos+,r);//右子树
    return root;
}
int main(){
    cin>>n;
    for(int i=;i<=n;i++) cin>>a[i];
    node *root=build(,n);//建树
    q.push(root);//bfs层序遍历
    while(!q.empty()){
        node *tree=q.front();
        q.pop();
        v.push_back(tree->d);
        if(tree->left) q.push(tree->left);
        if(tree->right) q.push(tree->right);
    }
    for(int i=;i<v.size();i++){//输出
        cout<<v.at(i);
        if(i!=v.size()-) cout<<" ";
    }
    return ;
}