PAT A 1064. Complete Binary Search Tree (30)

题目

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

即根据输入构建完全二叉树，并按层序输出。

一棵深度为n的满的完全二叉树元素数量为2^n-1；

一棵深度为n+1的不满的完全二叉树，即为深度为n的满的完全二叉树再加上深度为n+1的元素；

底层元素中的前2^(n-1)个元素小于根，之后的大于根。

根据这个性质可以求出一个排序区间中根的编号。

代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;

struct be	//探测结构，记录探测区域的起始点
{
int begin;
int end;
};

int Root(int p,int q);	//	求区域的根

int main()
{
int n;
vector<int> data;

int i,temp;
cin>>n;
for(i=0;i<n;i++)
{
cin>>temp;
data.push_back(temp);
}
sort(data.begin(),data.end());

queue<be> level;
int root;
be tempbe1,tempbe2;
tempbe1.begin=0;
tempbe1.end=n-1;
level.push(tempbe1);
while(!level.empty())
{
tempbe1=level.front();
level.pop();
root=Root(tempbe1.begin,tempbe1.end);
if(root-1>=tempbe1.begin)
{
tempbe2.begin=tempbe1.begin;
tempbe2.end=root-1;
level.push(tempbe2);
}
if(root+1<=tempbe1.end)
{
tempbe2.begin=root+1;
tempbe2.end=tempbe1.end;
level.push(tempbe2);
}
if(level.empty())
cout<<data[root];
else
cout<<data[root]<<" ";
}

return 0;
}

int Root(int p,int q)	//求根
{
if(p==q)
return p;
int d=q-p+1;	//区间元素数

int temp=1;
while(temp<d)	//获取一个容得下所有元素的树填满的容量
{
temp++;
temp*=2;
temp--;
}
temp++;	//求除去底层叶子后的元素数量
temp/=2;
temp--;

if(d-temp<=(temp+1)/2)	//底层元素数小于底层容量一半，即底层元素都小于根节点
return p+d-temp-1+(temp+1)/2;
else	//大于，则有一半比其小
return p+temp;
}