04-树6 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 $N$ ($\le 1000$). 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

参照别人代码，很妙

#include <stdio.h>  #include <stdlib.h>    int b[1005];  int j=0;    int compare(const void * a,  const void * b);  void mid_tre(int root,int N,int a[]);    int main(){      int N;      int i=0;      scanf("%d",&N);      int a[N];      for(i=0;i<N;i++){         scanf("%d",&a[i]);      }      qsort(a,N,sizeof(int),compare);      mid_tre(1,N,a);      printf("%d",b[1]);      for(i=2;i<=N;i++){          printf(" %d",b[i]);      }  }     int compare(const void * a, const void * b)   {       return *(int *)a - *(int *)b;   }      void mid_tre(int root,int N,int a[]){     if(root<=N){          mid_tre(2*root,N,a);          b[root]=a[j++];          mid_tre(2*root+1,N,a);       }   }