PAT 1064. Complete Binary Search Tree

Posted 2020-10-22 A-Little-Nut

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

分析

首先这是一个二叉搜索树，它的中序遍历结果时一个非递减的数列，并且又是个完全二叉树，就如果用数组去储存的话，已知父节点下标为s的话，左孩子的下标为2*s+1,右孩子的下标为2*s+2,并且输出的格式是层级输出，即一层一层的从左到右输出。这道题相当于已知中序遍历的结果（给的数组从小到大排序）了，让你输出层级输出的结果。
因为一旦完全二叉搜素树确定了，由于它的特定结构，它的层级输出的顺序也就确定了，关系是一一对应的，只需对0~n-1也中序遍历一遍，并和完全二叉树的中序遍历结果一一对应，再输出即可。

#include<iostream>
#include<algorithm>
using namespace std;
int a[1001],res[1001],tag=0;
void inorder_traversal(int s,int n){
     if(2*s+1<n) pre(2*s+1,n);
     res[s]=a[tag++];
     if(2*s+2<n) pre(2*s+2,n);
}
int main(){
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i];
    sort(a,a+n);
    inorder_traversal(0,n);
    for(int i=0;i<n;i++)
        i>0?cout<<" "<<res[i]:cout<<res[i];
    return 0;
}