【PAT甲级】1064 Complete Binary Search Tree（30 分）

题目链接
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.

A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.

Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding complete binary search 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:

10
1 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

题意:给定n个数构建完全二叉树，输出完全二叉树的层序遍历

思路：二叉树的中序遍历建树即为输出

代码：

#include 
using namespace std;
#define rep(i,a,n) for(int i=a;i=a;i--)
#define mem(a,n) memset(a,n,sizeof(a))
#define lowbit(i) ((i)&(-i))
typedef long long ll;
typedef unsigned long long ull;
const ll INF=0x3f3f3f3f;
const double eps = 1e-6;
const int N = 1e5+5;

int res[N],a[N];
int n,id=0;
void inOrder(int root) {
    if(root>=n) return ;
    inOrder(2*root+1);
    res[root]=a[id++];
    inOrder(2*root+2);
}
int main() {
    cin>>n;
    for(int i=0; i>a[i];
    }
    sort(a,a+n);
    inOrder(0);
    for(int i=0; i