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 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[], b[], k = ;
void inorder(int root)
{
    //cout<<"root:"<<root<<endl;
    if (root<n)
    {
        inorder( * root + );
        //printf("%d ==== %d ==== %d\n",root,k,a[k]);
        b[root] = a[k++];
        inorder( * root + );
    }
}

int main()
{
    cin >> n;
    for (int i = ; 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();
    for (int i = ; i<n - ; i++)
        cout << b[i] << " ";
    cout << b[n-] << endl;
    return ;
}

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