04-树6完整的二叉查找树 分数 30

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：

  二叉查找树（BST）被递归地定义为二叉树，该二叉树具有以下属性:

• 节点的左子树只包含键小于该节点键的节点。
• 节点右边的子树只包含键大于或等于节点键的节点。
• 左右子树也必须是二分搜索法树。

  完全二叉树（CBT）是一种完全填充的树，可能除了底层是从左到右填充的。

  现在给定一系列不同的非负整数键，如果要求树也必须是CBT，则可以构造唯一的BST。你应该输出这个BST的层次顺序遍历序列。

  输入规格:

  每个输入文件包含一个测试用例。对于每种情况，第一行包含一个正整数N (≤1000).然后N下一行给出了不同的非负整数键。一行中的所有数字由空格分隔，并且不大于2000。

  输出规格:

  对于每个测试用例，在一行中打印相应的完整二叉查找树的层次顺序遍历序列。一行中的所有数字必须用空格隔开，并且行尾不能有多余的空格。

思路：

将读入的n个数排序，然后计算当前根的左子树和右子树分别有多少节点，在当前[l,r]区间中，左子树是[l,l+lt-1]，右子树是[r-rt+1,r]，lt、rt是左、右子树节点数，根为mid = l+lt = r-rt。

#include <bits/stdc++.h>
using namespace std;
int n,flag = 0;
int a[1005];
int h[11] = {0,1,3,7,15,31,63,127,255,511,1023};
queue<int>q;
int f(int l,int r){
	if(l == r)return l;
	else{
		int cnt = 1;
		for(int i = 1;i<11;i++){
			if(r-l+1==h[i]){//求当前树的层高 
				cnt = i;
				break;
			}else if(r-l+1>h[i]){
				cnt++;
			}else break;
		}
		int len = r-l+1 - h[cnt-1];//求最下面一层的节点数 
		int llen,rlen;			//最下面一层节点中属于左子树的数量和属于右子树的数量 
		if(len<=h[cnt-2]+1){	
			llen = len;
			rlen = 0;
		}else{
			llen = h[cnt-2]+1;
			rlen = len - llen;
		}
		int lt = llen + (r-l-len)/2;		//求得左右子树的节点个树 
		int rt = rlen + (r-l-len)/2;
		int mid = l + lt;
		if(mid-1>=l){
			q.push(l);
			q.push(mid-1);
		}
		if(r>=mid+1){
			q.push(mid+1);
			q.push(r);
		}
		return mid;
	}
}
int main()
{
	cin>>n;
	for(int i = 1;i<=n;i++){
		cin>>a[i];
	}
	sort(a+1,a+1+n);
	q.push(1);
	q.push(n);
	while(!q.empty()){
		int l,r;
		l = q.front();
		q.pop();
		r = q.front();
		q.pop();
		int mid = f(l,r);
		if(flag)cout<<" ";
		else flag = 1;
		cout<<a[mid];
	}
	return 0;
}