PAT 甲级 1064 Complete Binary Search Tree (30 分)（不会做，重点复习，模拟中序遍历）...
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
题意：
给出二叉搜索树的一个序列，如果这棵树还要求是完全二叉树的话，这棵树就是唯一的。现在要输出这棵树的层序遍历。
题解:
看了别人的代码，也不是很能理解。。。。
一棵排序二叉树的中序遍历就是这一组数的递增序列。
这边是完全二叉树，假设从0开始，那么节点i的左孩子的标号就是2*i+1，右孩子的标号就是2*(i+1)。
先将这组数按照递增来排序，然后用中序遍历复原这棵完全排序二叉树，最后直接输出。
完全二叉排序树按层序存放在从1开始的数组中，左右儿子节点的索引分别为（根节点索引假设为root）root*2和root*2+1。而二叉排序树的中序遍历是递增的。我们现在把输入序列排序，就可以得到中序遍历的序列了。于是我们开始遍历这棵左右儿子节点的索引分别为（根节点索引假设为root）root*2和root*2+1的空树，与以往一边遍历一边打印不同的是，我们是一边遍历，一边给这棵空树赋值。
很不错。学习这种思想。
AC代码：
#include<iostream>
#include<set>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
int n, a[1005], b[1005], k = 0;
void inorder(int root)
{
//cout<<"root:"<<root<<endl;
if (root<n)
{
inorder(2 * root + 1);
//printf("%d ==== %d ==== %d\n",root,k,a[k]);
b[root] = a[k++];
inorder(2 * root + 2);
}
}
int main()
{
cin >> n;
for (int i = 0; i<n; i++)
cin >> a[i];
sort(a, a + n); //从小到大排序 因为完全二叉搜索树的中序遍历就是从小到大排序
inorder(0);
for (int i = 0; i<n - 1; i++)
cout << b[i] << " ";
cout << b[n-1] << endl;
return 0;
}
10
1 2 3 4 5 6 7 8 9 0
root:0
root:1
root:3
root:7
root:15
7 ==== 0 ==== 0
root:16
3 ==== 1 ==== 1
root:8
root:17
8 ==== 2 ==== 2
root:18
1 ==== 3 ==== 3
root:4
root:9
root:19
9 ==== 4 ==== 4
root:20
4 ==== 5 ==== 5
root:10
0 ==== 6 ==== 6
root:2
root:5
root:11
5 ==== 7 ==== 7
root:12
2 ==== 8 ==== 8
root:6
root:13
6 ==== 9 ==== 9
root:14
6 3 8 1 5 7 9 0 2 4
