PAT 甲级 1064 Complete Binary Search Tree (30 分)（不会做，重点复习，模拟中序遍历）

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

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
",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