【题解】【PAT甲】1064 Complete Binary Search Tree（30分）（BST）（完全二叉树）（中序转层序）

题目链接

 PTA | 程序设计类实验辅助教学平台

题目描述

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

结尾无空行

Sample Output:

6 3 8 1 5 7 9 0 2 4

结尾无空行

题目大意

 给你一些节点，让你构成完全搜索二叉树，并输出层序序列

解题思路

 因为要是搜索二叉树，所以中序必定是从小到大顺序排序的，所以只要将给定节点的值sort排序，然后就可以得到中序遍历的值，然后利用中序转层序的方法即可得到答案

void inOrder(int root){
	if(root>=n)	return ;
	inOrder(root*2+1);
	level[root]=in[t++];
	inOrder(root*2+2);
}

题解

#include<bits/stdc++.h>
using namespace std;
vector<int> in;
int n;
vector<int> level;
int t=0;
void inOrder(int root){
	if(root>=n)	return ;
	inOrder(root*2+1);
	level[root]=in[t++];
	inOrder(root*2+2);
}
int main(){
	cin>>n;
	in.resize(n);
	level.resize(n);
	for(int i=0;i<n;i++){
		cin>>in[i];
	}
	sort(in.begin(),in.end());
	inOrder(0);
	for(int i=0;i<level.size();i++){
		if(i!=0)	cout<<" ";
		cout<<level[i];
	}
}