pat 甲级 1064 ( Complete Binary Search Tree ) (数据结构)

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 (≤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

重点在递归的过程建树

详见代码：

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
int a[1005];
int cnt[12]={-1,1,3,7,15,31,63,127,255,511,1023};
struct node
{
    int v;
    node *left=NULL,*right=NULL;
};
int mypow(int a,int b)
{
    int res=1;
    while(b--)res*=a;
    return res;
}
node* build(int l,int r,int num)
{
    if(l>r){return NULL;}
    if(l==r)
    {
        node *tt=new node();tt->v=a[l];
        return tt;
    }
    if(num<=0)return NULL;
    node *thi;
    int k1=0;
    while(cnt[k1+1]<num)k1++;
    int k2=k1+1;
    int rr=num-(cnt[k1]);
    int tmp=mypow(2,k2-2);
    if(rr>=tmp)
    {
        int lnum=cnt[k1];
        int rnum=num-lnum-1;
        int id=l+lnum;
        thi=new node();
        thi->v=a[id];
        thi->left=build(l,id-1,lnum);
        thi->right=build(id+1,r,rnum);
    }
    else {
        int lnum=num-1-cnt[k1-1];
        int rnum=cnt[k1-1];
        int id=l+lnum;
        thi=new node();thi->v=a[id];
        thi->left=build(l,id-1,lnum);
        thi->right=build(id+1,r,rnum);
    }
    return thi;
}
int main()
{
    //freopen("in.txt","r",stdin);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    sort(a+1,a+1+n);
    int l=1;int r=n;
    node *head=build(l,r,n);
    queue< node* >q;
    while(!q.empty())q.pop();
    q.push(head);
    vector<int>ans;ans.clear();
    node *now;
    while(!q.empty())
    {
        now=q.front();q.pop();
        ans.push_back(now->v);
        if(now->left!=NULL)q.push(now->left);
        if(now->right!=NULL)q.push(now->right);
    }
    for(int i=0;i<ans.size();i++)
    {
        cout<<ans[i];if(i!=ans.size()-1)cout<<" ";
    }
    cout<<endl;
    return 0;
}