甲级PAT 1064 Complete Binary Search Tree （30 分）(完全二叉查找树，BST,CBT）
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题目要求
给出一个序列，构建完全二叉查找树，输出其层序遍历的结果
解题思路
二叉查找树中序遍历所得的结果是有序的，按照从小到大排序。
完全二叉树对于给定数量的结点其形状固定。
根据这一性质，对一个给结点数n，就可以确定其根结点在有序序列里的下标。
首先将序列a按从小到大排序，对于10个结点，可以根据序列的下标构造一棵如下图的完全二叉查找树。
可以看出，根据这棵完全二叉树很容易就能写出其层序遍历。
我们主要的任务就是找到每个结点对应的在有序序列中的下标。
利用递归实现。
对于10个结点，其根结点7，是由一个基址（因为右子树的基址和左子树的基址不同，左子树基址0，右子树基址父节点下标）+其左子树结点的个数+1得到的，
这里基址0，左子树结点个数6，所以0+6+1=7
层序遍历数组下标1的值就为level[1]=a[7]
然后需要分别对根结点7的左子树和右子树进行处理，由于层序遍历，是左子树再右子树前面，左子树根结点的层序下标为2*1,右子树根结点的下标为2*1+1。
左子树结点的个数是根据结点的个数所能构建的完全二叉树的层数height，height从1开始，1层结点总个数2^1-1,2层结点总个数2^2-1，第1层为根结点，对于n个结点的左子树的个数有两种可能情况
所以左子树的个数为min(pow(2,height-1)-1,n-pow(2,height-2))
参考代码
#include
using namespace std;
int level[1010],a[1010];
int findleftnum(int n){
int height=1;
while(pow(2,height)-1>n;
for(i=1;i<=n;i++) cin>>a[i];
sort(a+1,a+n+1);
create(0,n,1);
for(i=1;i<=n;i++){
if(i!=1) cout<<" ";
cout<