PAT真题练习（甲级）1064 Complete Binary Search Tree (30 分)

PAT真题练习（甲级）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 (≤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 1:

10
1 2 3 4 5 6 7 8 9 0

Sample Output 1:

6 3 8 1 5 7 9 0 2 4

一点想法

这道题主要是对完全二叉搜索树性质的考察
首先，我们要知道，完全二叉搜索树的中序遍历结果，为所有数字从小到大的排列。
其次，我们常用一个数组来表示完全二叉搜索树，并满足对于root节点 root * 2节点为他的左子，root * 2 + 1为他的右子
因此，本题中，我们读入数据后，对其进行排序，获得该树中序遍历的结果。之后对树进行中序遍历，并依照顺序赋值（类比于中序遍历输出结果，这里只是把输出结果改成赋值）。

AC代码

#include
#include 
#include 
#include 
#include 
#include 
using namespace std;

vector<int> tree;
vector<int> numbers;
int id = 0;
int N;

// 对树进行中序遍历，即可按顺序赋值
void inorder(int root) {
	if (root > N) return;
	inorder(root * 2);
	tree[root] = numbers[id++];
	inorder(root * 2 + 1);
}

int main() {
	cin >> N;
	numbers.resize(N);
	tree.resize(N + 1);
	for (auto i = 0; i < N;i++) {
		cin >> numbers[i];
	}
	// 排序获得中序遍历的结果
	sort(numbers.begin(), numbers.end());
	inorder(1);
	for (auto i = 1;i < N;i++) {
		cout << tree[i] << " ";
	}
	cout << tree[N];

}