1064 Complete Binary Search Tree (30 分)

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。

#include <bits/stdc++.h>

using namespace std;
int n;
int an[3000];
struct Node
{
    int left, right;
};
int val[11] = {1,2,4,8,16,32,64,128,256,512,1024};
queue<Node> q;
vector<int> v;
int main(){
    cin >> n;
    for(int i = 0 ; i < n; i++){
        cin >> an[i];
    }
    sort(an, an+n);
    q.push({0,n-1});
    while(!q.empty()){
        Node node = q.front();
        q.pop();
        int ll = node.left, rr = node.right;
        int len = rr-ll+1;
        int pos = 0;
        while(val[pos] <= len){
            len -= val[pos];
            pos ++;
        }
        int ans = val[pos]/2 + min(val[pos]/2, len);
        v.push_back(an[ans+ll-1]);
        if(ll <= ans+ll-2){
            q.push({ll, ans+ll-2});
        }
        if(ans+ll <= rr){
            q.push({ans+ll, rr});
        }
    }
    for(int i = 0 ; i < v.size(); i++){
        printf("%d%c", v[i], i == v.size()-1?'\n':' ');
    }
    return 0;
}