PAT Advanced 1064 Complete Binary Search Tree (30) [⼆叉查找树BST]

题目

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The lef 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 lef 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 lef 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

题目分析

已知完全二叉查找树的序列，求其层序序列

解题思路

思路 01

1. 输入测试用例，升序排序即为二叉查找树的中序序列
2. 中序序列递归建树（存储与数组i节点的子节点为2*i,2*i+1，root在1位置）（模拟递归中序遍历过程）
3. 顺序打印数组即为层序序列

思路 02

1. 输入测试用例，升序排序即为二叉查找树的中序序列
2. 中序序列递归建树（存储与数组i节点的子节点为2i,2i+1，root在1位置）（找root的坐标k，将中序序列划分为start_{k-1左子树和k+1}end右子树）
3. 顺序打印数组即为层序序列

知识点

完全二叉查找树，任意节点序列建树
注：二叉查找树（不一定要完全二叉树）任意节点序列，升序排序，即为二叉查找树的中序序列

1. 方式一：模拟递归打印中序序列的过程，将中序序列依次插入到数组建树
2. 方式二：找root在中序序列中的位置k并将root保存到数组，递归处理start_{k-1左子树和k+1}end右子树，完成建树

void inOrder(int root) { //root保存在1位置，index初始化为0
	if(root>n) return;
	inOrder(root*2);
	CBT[root]=number[index++];
	inOrder(root*2+1);
}

void getLevel(int start, int end, int index) {
	if(start>end)return;
	int n=end-start+1;
	int l=log(n+1)/log(2);//最后一层的层数
	int leave=n-(pow(2,l)-1); //最后一层叶子结点数
	//pow(2, l - 1) - 1是除了root结点所在层和最后?层外，左?树的结点个数，pow(2, l - 1) 是l+1
	//层最多拥有的属于根结点左?树的结点个数，min(pow(2, l - 1), leave)是最后?个结点真正拥有的
	//属于根结点左?树上的结点个数
	int root = start+(pow(2,l-1)-1)+min((int)pow(2,l-1),leave);
	level[index]=in[root];
	getLevel(start,root-1,2*index+1);
	getLevel(root+1,end,2*index+2);

}

Code

Code 01

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=1000;
int n,m;
int number[maxn],CBT[maxn],index=0;
void inOrder(int root) {
	if(root>n) {
		return;
	}
	inOrder(root*2);
	CBT[root]=number[index++];
	inOrder(root*2+1);
}
int main(int argc,char * argv[]) {

	scanf("%d",&n);
	for(int i=0; i<n; i++) {
		scanf("%d",&m);
		number[i]=m;
	}
	// 递增排序--二叉查找树中序序列
	sort(number,number+n);
	inOrder(1);//根节点在1位置
	for(int i=1; i<=n;i++) {
		if(i!=1)printf(" ");
		printf("%d",CBT[i]);
	}
	return 0;
}

Code 02

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
vector<int> in,level;
void getLevel(int start, int end, int index) {
	if(start>end)return;
	int n=end-start+1;
	int l=log(n+1)/log(2);//最后一层的层数
	int leave=n-(pow(2,l)-1); //最后一层叶子结点数
	//pow(2, l - 1) - 1是除了root结点所在层和最后?层外，左?树的结点个数，pow(2, l - 1) 是l+1
	//层最多拥有的属于根结点左?树的结点个数，min(pow(2, l - 1), leave)是最后?个结点真正拥有的
	//属于根结点左?树上的结点个数
	int root = start+(pow(2,l-1)-1)+min((int)pow(2,l-1),leave);
	level[index]=in[root];
	getLevel(start,root-1,2*index+1);
	getLevel(root+1,end,2*index+2);

}
int main(int argc,char * argv[]) {
	int n;
	scanf("%d",&n);
	in.resize(n);
	level.resize(n);
	for(int i=0; i<n; i++) {
		scanf("%d",&in[i]);
	}
	sort(in.begin(),in.end());
	getLevel(0,n-1,0);
	printf("%d",level[0]);
	for(int i=1; i<n; i++) {
		printf(" %d",level[i]);
	}
	return 0;
}