pat 1064

1064. Complete Binary Search Tree (30)

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

101 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>#include <queue>using namespace std;typedef struct Tree{Tree *lchild;Tree *rchild;int data;} Tree, *Llink;Tree tree[1010];int n, loc = 0;int order[1010], lvOd[1010], sum;queue<Llink> Q;Tree *creat(){tree[loc].lchild = tree[loc].rchild = NULL;return &tree[loc++];}//建立完全二叉搜索树，关键是每次递归时找到根结点的位置，和左子树结点的个数和右子树结点的个数Tree *build(int s, int e){Tree *T = NULL;if(s > e)return T;int num = e - s + 1, i = 0, level, remain;while(num >= (int)pow(2.0, i)){num -= pow(2.0, i);i++;}level = i;remain = num;//最后一层结点的个数int countl =(e - s - remain) / 2;int countr = countl; if( (int)pow(2.0, level) / 2 >= num)countl += num;else{countl += (int)pow(2.0, level) / 2; //左子树结点的个数countr += num - (int)pow(2.0, level) / 2; //右子树结点的个数}T = creat();T->data = order[s+countl];  //根结点的位置T->lchild = build(s, s + countl - 1);T->rchild = build(s + countl + 1, e);return T;}void levelTraversal(){int i;Tree *T = NULL;while(!Q.empty()){T = Q.front();Q.pop();lvOd[sum++] = T->data;if(T->lchild != NULL)Q.push(T->lchild);if(T->rchild != NULL)Q.push(T->rchild);}}int main(){int i, j;while(scanf("%d", &n) != EOF){Tree *T = NULL;for(i = 0; i < n; i++){scanf("%d", &order[i]);}sort(order, order + n);T = build(0, n-1);while(!Q.empty())Q.pop();sum = 0;Q.push(T);levelTraversal();for(i = 0; i < sum; i++){if(i != sum - 1)printf("%d ", lvOd[i]);elseprintf("%d\n", lvOd[i]);}}return 0;}