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:

  101 2 3 4 5 6 7 8 9 0

  Sample 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
       
        #define MAXLEN 2000int 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
        
         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\n");//	Outputraw();//	printf("\n");	CountNodes();	CalcTree(workarray,CBST,0,1);	Output();	printf("\n");//	Outputcount();}