1064 Complete Binary Search Tree (30 分)  

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 (≤). 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


这个二叉树挺好的，我用的是bfs遍历写的。
我看到其他的博客都是建树，我比较懒（逃。。。）
算是分治的写法了。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 int n;
 5 int an[3000];
 6 struct Node
 7 {
 8     int left, right;
 9 };
10 int val[11] = {1,2,4,8,16,32,64,128,256,512,1024};
11 queue<Node> q;
12 vector<int> v;
13 int main(){
14     cin >> n;
15     for(int i = 0 ; i < n; i++){
16         cin >> an[i];
17     }
18     sort(an, an+n);
19     q.push({0,n-1});
20     while(!q.empty()){
21         Node node = q.front();
22         q.pop();
23         int ll = node.left, rr = node.right;
24         int len = rr-ll+1;
25         int pos = 0;
26         while(val[pos] <= len){
27             len -= val[pos];
28             pos ++;
29         }
30         int ans = val[pos]/2 + min(val[pos]/2, len);
31         v.push_back(an[ans+ll-1]);
32         if(ll <= ans+ll-2){
33             q.push({ll, ans+ll-2});
34         }
35         if(ans+ll <= rr){
36             q.push({ans+ll, rr});
37         }
38     }
39     for(int i = 0 ; i < v.size(); i++){
40         printf("%d%c", v[i], i == v.size()-1?'\n':' ');
41     }
42     return 0;
43 }