《数据结构》04-树6 Complete Binary Search Tree

我会带着你远行 2022-04-22 05:52 125阅读 0赞

# 题目 #

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 #

大概题意是：给定一组二叉搜索树插入值，使得刚好该二叉搜索树是棵完全二叉树，输出组成该树的值的层序遍历

考虑到链表的层序遍历还得单独实现，采用数组实现二叉搜索树  
整体框架是一个先序遍历，由于给了一组输入，排好序就成了该二叉搜索树的中序序列，利用完全二叉树的性质可以求出左子树结点个数，从而递归填充根结点

#include<iostream>
    #include<algorithm>
    #include<cmath>
    #include<vector>
    #define MaxSize 2005
    using namespace std;
    int value[MaxSize];
    int BST[MaxSize];
    
    // 计算 n 个结点的树其左树结点个数 
    int getLeftTreeSize(int n){
        
    	int h =0;   // 保存该结点下满二叉树的层数 
    	int tmp = n+1;
    	while(tmp!=1){
        
    		tmp /=2;
    		h++;
    	}
    	int x = n-pow(2,h)+1;   // 最下面一排子叶结点个数 
    	x = x<pow(2,h-1)?x:pow(2,h-1);   // 子叶结点个数最多是 2^(h-1) 
    	int L = pow(2,h-1)-1+x;   // 该结点个数情况下左子树的个数 
    	return L;
    }
    
    // 填充函数
    void fill(int left,int right,int root){
        
    	int n = right - left + 1;  // 确定范围内数值个数 
     	if(!n)
     		return;
     	int L = getLeftTreeSize(n);   // 找到"偏移量" 
     	BST[root] = value[left + L];    // 根结点的值应该是 左边界值 + 偏移量 
     	int leftRoot = 2 * root + 1;   // 左儿子结点位置，由于从 0 开始 
     	int rightRoot = leftRoot + 1;  // 右儿子结点位置 
     	fill(left,left+L-1,leftRoot);  // 左子树递归 
     	fill(left+L+1,right,rightRoot);   // 右子树递归 
    }
    
    int main(){
        
    	int n;
    	cin>>n;
    	for(int i=0;i<n;i++){
        
    		cin>>value[i];
    	}
    	// 从小到大排序 
    	sort(value,value+n);
    	fill(0,n-1,0);
    	for(int i=0;i<n;i++){
        
    		if(i)
    			cout<<" ";
    		cout<<BST[i];
    	}
    	return 0;
    }