pat 甲级 1064 ( Complete Binary Search Tree ) (数据结构)

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本文探讨了如何从一组非负整数键中构建一个既为二叉搜索树又为完全二叉树的数据结构。通过递归过程，文章详细解释了如何确保树的左子树仅包含小于节点键的值，右子树则包含大于等于节点键的值，并且树的底部层级从左到右填充。最后，给出了一个具体示例，展示了构建的完全二叉搜索树的层次遍历序列。

1064 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 (≤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

重点在递归的过程建树

详见代码：

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
int a[1005];
int cnt[12]={-1,1,3,7,15,31,63,127,255,511,1023};
struct node
{
    int v;
    node *left=NULL,*right=NULL;
};
int mypow(int a,int b)
{
    int res=1;
    while(b--)res*=a;
    return res;
}
node* build(int l,int r,int num)
{
    if(l>r){return NULL;}
    if(l==r)
    {
        node *tt=new node();tt->v=a[l];
        return tt;
    }
    if(num<=0)return NULL;
    node *thi;
    int k1=0;
    while(cnt[k1+1]<num)k1++;
    int k2=k1+1;
    int rr=num-(cnt[k1]);
    int tmp=mypow(2,k2-2);
    if(rr>=tmp)
    {
        int lnum=cnt[k1];
        int rnum=num-lnum-1;
        int id=l+lnum;
        thi=new node();
        thi->v=a[id];
        thi->left=build(l,id-1,lnum);
        thi->right=build(id+1,r,rnum);
    }
    else {
        int lnum=num-1-cnt[k1-1];
        int rnum=cnt[k1-1];
        int id=l+lnum;
        thi=new node();thi->v=a[id];
        thi->left=build(l,id-1,lnum);
        thi->right=build(id+1,r,rnum);
    }
    return thi;
}
int main()
{
    //freopen("in.txt","r",stdin);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    sort(a+1,a+1+n);
    int l=1;int r=n;
    node *head=build(l,r,n);
    queue< node* >q;
    while(!q.empty())q.pop();
    q.push(head);
    vector<int>ans;ans.clear();
    node *now;
    while(!q.empty())
    {
        now=q.front();q.pop();
        ans.push_back(now->v);
        if(now->left!=NULL)q.push(now->left);
        if(now->right!=NULL)q.push(now->right);
    }
    for(int i=0;i<ans.size();i++)
    {
        cout<<ans[i];if(i!=ans.size()-1)cout<<" ";
    }
    cout<<endl;
    return 0;
}