1064 Complete Binary Search Tree （运用搜索二叉树BST的中序遍历输出层序遍历）

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 (≤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

参考大佬博客：https://blog.csdn.net/richenyunqi/article/details/78868563

本题思路

方法一：利用完全搜索二叉树的性质

n为结点个数

1.先将输入的n个结点值排序得到中序排列，

2. 通过log(n + 1) 向下取整 (floor() 为向下取整， ceil为向上取整) 得到除去最后一层的高度h eg:n=7的结点高度为3 （因为根节点只有一个值所以需要 log n+1 除去最后一层高度）

3.得到最后一层的结点数： n - (pow(2, h) - 1); (可能只存在左结点，也存在右结点)

4. 在测试样例1 中序排列中： 0 1 2 3 4 5 6 7 8 9 只要知道了左结点的个数 num; 通过(left(中序遍历结点最左边) + num 就能确定 root结点,)找到了根结点后然后不断递归先左子树，后右子树，设定一个index 记录根的下标方便层序输出

使 temp2 = (pow(2,h) - 1 ) / 2

5. 关键就如何求出左子树结点个数：此时 temp2 得到最后一层左子树能够存放的最大结点数(最后一层能够存储最大容量的一半)，通过min(temp2 , n - (pow(2, h) - 1) ）得到此时最后一层能够存放的左子树元素， temp2 -1 可以得到除去最后一层的左子树个数：因此左结点个数和sum为： sum =temp2 - 1 + min(temp2 , n - (pow(2, h) - 1) ）

具体代码

：

#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
int n;
vector<int> in,level;
void levelOrder(int start, int end, int index){
	if(start > end) return;
    int sum = end - start + 1;
	int height = (int)floor(log((sum) + 1) / log(2)); //还没强转 log在cmath库中 
	int temp1 = sum - (int)(pow(2, height) - 1);  //最后一层的结点数量 
	int temp2 = pow(2, height) / 2; 	   //最后一层能够容纳的最大结点数量的一半 
	temp1 = min(temp1, temp2);         //取最小是因为temp1可能包含右结点，现在只需要算出左结点
	temp1 = start + temp1 + temp2 - 1;         //此时左子树结点总数为（temp1 + temp2 - 1） 同时加上start(left)就是中序遍历中根结点的下标
	level[index] = in[temp1];
	levelOrder(start, temp1 - 1, 2 * index + 1);
	levelOrder(temp1 + 1, end, 2 * index + 2);
}	
int main(){
	scanf("%d", &n);
	in.resize(n);    //在全局遍历中 用（n）也不能scanf输入 这里注意 只有用resize容器具有具体长度时 才能用scanf赋值 
	level.resize(n);
	for(int i = 0; i < n; i++){
		scanf("%d", &in[i]);
	}
	sort(in.begin(), in.end());
	levelOrder(0, n-1, 0);	
	for(auto i = 0; i < n; i++){
		if(i != 0) printf(" ");
		printf("%d", level[i]);
	}
	return 0;
}

方法二：

方法一是通过顺序查找到每个根结点将每个结点的下标索引放入level数组中，

更简便的方法是直接通过中序遍历将中序遍历中的每个元素都加入到level数组中并不需通过求出高度，找到左结点的个数先序遍历找到每个根结点

具体代码：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int n,position=0;
vector<int> level, in;
void levelOrder(int index){
	if(index > n-1) return;      //注意这里是n-1 因为根节点的下标是0 那么最后一个结点的下标为n-1 
	levelOrder(index * 2 + 1);   // 完全二叉搜索树中中序遍历中 第一个结点会是最左边的结点 
	level[index] = in[position++];
	levelOrder(index * 2 + 2);
}
int main(){
	scanf("%d", &n);
	level.resize(n);in.resize(n);
	for(int i = 0; i < n; i++){
		scanf("%d", &in[i]);
	}
	sort(in.begin(),in.end());
	levelOrder(0);
	for(int i = 0; i < n; i++){
		if(i != 0) printf(" ");
		printf("%d", level[i]);
	}
	return 0;
}