题目地址
https://pta.patest.cn/pta/test/16/exam/4/question/669
5-7 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 NN (le 1000≤1000). Then NN 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 0Sample Output:
6 3 8 1 5 7 9 0 2 4
/*
评测结果
时间 结果 得分 题目 编译器 用时（ms） 内存（MB） 用户
2017-06-29 10:45 正在评测 0 5-7 gcc 无 无
测试点结果
测试点 结果 得分/满分 用时（ms） 内存（MB）
测试点1 答案正确 18/18 1 1
测试点2 答案正确 3/3 2 1
测试点3 答案正确 2/2 1 1
测试点4 答案正确 2/2 1 1
测试点5 答案正确 3/3 1 1
测试点6 答案正确 2/2 3 1
完全二叉搜索树
结构：数组形式存储完全树，可以很方便进行层序遍历
解法：思考后发现完全二叉搜索树的子树也是完全二叉搜索树。故先将接收到的数据排序，递归计算一个节点的左子树和右子树数量，类似于快排，每次可以确定root的值。
然后按顺序打印数组即可。
*/
#include<stdio.h>
#define MAXLEN 2000
int workarray[MAXLEN];
int CBST[MAXLEN+1];//0号元素空着，从1号开始左子树2*root，右子树2*root+1；
int NODECOUNT[MAXLEN+1];
int Len;
void swap(int *a,int *b)
{
int temp;
temp=*a;
*a=*b;
*b=temp;
}
void InsertionSort(int A[],int N)
{
int i;
int k;
for(i=1;i<N;i++)
{
int temp=A[i];
k=i;
while(temp < A[k-1])
{
swap(&A[k],&A[k-1]);
k--;
}
A[k]=temp;
}
}
void GetInput()
{
int i;
scanf("%d",&Len);
for(i=0;i<Len;i++)
{
scanf("%d",&workarray[i]);
}
}
void Output()
{
int i;
for(i=1;i<Len+1;i++)
{
printf("%d",CBST[i]);
if(i!=Len)
printf(" ");
}
}
void Outputcount() //调试用函数
{
int i;
for(i=1;i<Len+1;i++)
{
printf("%d ",NODECOUNT[i]);
}
}
void Outputraw() //调试用函数
{
int i;
for(i=0;i<Len;i++)
{
printf("%d ",workarray[i]);
}
}
void CountNodes() //太懒了，数据量不大，直接让它自己数节点数量得了。 逆序算出此节点之下所有子树节点之和
{
int i;
for(i=0;i<Len+1;i++)
{
NODECOUNT[i]=0;//先全体置零
}
for(i=Len;i>0;i--)//在1开头的数组中，有效下标范围应该是1~Len
{
NODECOUNT[i/2] += NODECOUNT[i]+1; //父节点的子树，加上本节点的子树，再加上该节点自身
}
}
void CalcTree(int raw[],int calc[],int left,int index)
{
if (index>Len) return; //越界了，没子树
int raw_position=left+(2*index>Len?0:NODECOUNT[2*index]+1);//注，函数首次调用，left最初传入的是0， 判断是否为叶节点，如果没有左子树，那么它自己就是left
//此处加个注释说明下，一个坐标的构成。left(此递归段偏移起点)+NODECOUNT[2*index](所求的点的左儿子的全部结点数量)+1(该结点本身，即此递归段的root)
calc[index]=raw[raw_position];
CalcTree(raw,calc,left,2*index);//递归左子树
CalcTree(raw,calc,raw_position+1,2*index+1);//递归右子树
}
int main()
{
GetInput();
InsertionSort(workarray,Len);
// printf("done sort
");
// Outputraw();
// printf("
");
CountNodes();
CalcTree(workarray,CBST,0,1);
Output();
printf("
");
// Outputcount();
}