1064. Complete Binary Search Tree (30)-PAT

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本文介绍了一种根据给定的非负整数序列构造完全二叉搜索树的方法，并通过示例展示了如何进行层次遍历输出。文章还提供了完整的C++实现代码。

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:

10
1 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

树的性质：高度，每层节点的个数

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<iostream>
#include<queue>
#include<math.h>
using namespace std;
int compare(const void *a,const void *b){
  return *(int *)a-*(int*)b;
}
typedef struct Node{
  int val;
  Node *left;
  Node *right;
}Node;
void build_tree(int *num,int len,Node **r){
  if(len>0){
    int i,high;
    for(i=0;i<1000;i++){
      if((int)pow(2.0,i)-1>len){
          high=i;
          break;
      }
    }
    int ct=(int)pow(2.0,high-1)-1;
    int remain=len-ct;
    int last_need=(int)pow(2.0,high-1);
    int left,right;
    if(remain>last_need/2)
      left=(ct-1)/2+last_need/2;
    else
      left=(ct-1)/2+remain;
    right=len-left-1;
      Node * root= new Node;
    *r=root;
    root->left=NULL;
    root->right=NULL;
    root->val=num[left];
    if(left>0)
      build_tree(num,left,&root->left);
    if(right>0)
      build_tree(num+left+1,right,&root->right);
  }
}
void print_tree(Node *root){
  if(root!=NULL){
      queue<Node*> q;
      q.push(root);
      printf("%d",root->val);
      while(!q.empty()){
        Node* node=q.front();
        if(node!=root)
          printf(" %d",node->val);
        if(node->left!=NULL)
          q.push(node->left);
        if(node->right!=NULL)
          q.push(node->right);
        q.pop();
      }
  }
}
int main()
{
  int n,i;
  scanf("%d",&n);
  int *num=new int[n];
  for(i=0;i<n;i++){
    scanf("%d",&num[i]);
  }
  qsort(num,n,sizeof(int ),compare);
  Node *root=NULL;
  build_tree(num,n,&root);
  print_tree(root);
  return 0;
}