pat04-树8. Complete Binary Search Tree (30)

04-树8. Complete Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

8000 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:

10
1 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

---

提交代码

解法一：

数组做法：

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<queue>
 #include<vector>
 #include<cmath>
 #include<string>
 using namespace std;
 int tree[],endtree[];
 int GetLeftLength(int n){
     int h=log(n+)/log();//除去最后一排的元素
     int x=n+-pow(,h);
     if(x>=pow(,h-)){
         x=pow(,h-);
     }
     return x+pow(,h-)-;
 }
 void solve(int left,int right,int root){
     int n=right-left+;
     if(!n) return;
     int l=GetLeftLength(n);
     endtree[root]=tree[left+l];
     solve(left,left+l-,root*+);
     solve(left+l+,right,root*+);
 }
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n;
     scanf("%d",&n);
     int i;
     for(i=;i<n;i++){
         scanf("%d",&tree[i]);
     }
     sort(tree,tree+n);
     solve(,n-,);
     printf("%d",endtree[]);
     for(i=;i<n;i++){
         printf(" %d",endtree[i]);
     }
     printf("\n");
     return ;
 }

方法二：

链表做法：

递归建树的思想：想找出当前的中间大的树，作为当前树根，然后遍历左子树，右子树，最后对所建的BST进行层序遍历。

当前元素总个数为n，则层数k为ceil(log2(n+1))：

1.n>=3*2^(k-2)，即查找树的最后一个元素位于根的右子树部分。则此时右子树有 n-(3*2^(k-2)-1)+2^(k-2)-1 个元素，左边有 n-右子树元素个数-1=n-(n-(3*2^(k-2)-1)+2^(k-2)-1)-1=2^(k-1)-1个元素，则中间元素下标为 2^(k-1)-1 （从0开始）。

2.n<3*2^(k-2)，即查找树的最后一个元素位于根的左子树部分。则此时右子树有 2^(k-2)-1 个元素，左边有 n-右子树元素个数-1=n-(2^(k-2)-1)-1=n-2^(k-2) 个元素，则中间元素下标为 n-2^(k-2) （从0开始）。

 #include<cstdio>
 #include<functional>
 #include<queue>
 #include<cmath>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 int mem[];
 struct node{
     int v;
     node *l,*r;
     node(){
         l=r=NULL;
     }
 };
 void CBTBuild(int *mem,int n,int mid,node *&p){
     //q.push(mem[mid]);
     p=new node();
     p->v=mem[mid];
     //cout<<mem[mid]<<endl;

     if(mid->=){//left tree
         int nn=mid;
         int k=ceil(log(nn+)/log());
         if(nn>=*pow(,k-)){//超过一半
             CBTBuild(mem,nn,pow(,k-)-,p->l);
         }
         else{//未超过一半
             CBTBuild(mem,nn,nn-pow(,k-),p->l);
         }
     }
     if(mid+<n){//right tree
         int nn=n-(mid+);
         int k=ceil(log(nn+)/log());
         if(nn>=*pow(,k-)){//超过一半
             CBTBuild(mem+mid+,nn,pow(,k-)-,p->r);
         }
         else{//未超过一半
             CBTBuild(mem+mid+,nn,nn-pow(,k-),p->r);
         }
     }
 }
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n,i,j,k;
     queue<node> q;
     scanf("%d",&n);
     for(i=;i<n;i++){
         scanf("%d",&mem[i]);
     }
     sort(mem,mem+n);

     /*for(i=0;i<n;i++){
         cout<<mem[i]<<endl;
     }*/
     node *h;
     k=ceil(log(n+)/log());
     if(n>=*pow(,k-)){//超过一半
         CBTBuild(mem,n,pow(,k-)-,h);

         //cout<<mem[int(pow(2,k-1)-1)]<<endl;

     }
     else{//未超过一半
         CBTBuild(mem,n,n-pow(,k-),h);

         //cout<<mem[int(n-pow(2,k-2))]<<endl;

     }
     int top;
     node p=*h;
     q.push(p);
     printf("%d",p.v);
     while(!q.empty()){
         p=q.front();
         q.pop();
         if(p.l!=NULL){
             q.push(*(p.l));
             printf(" %d",p.l->v);
         }
         if(p.r!=NULL){
             q.push(*(p.r));
             printf(" %d",p.r->v);
         }
     }
     printf("\n");
     return ;
 }