1064. Complete Binary Search Tree (30)
时间限制
100 ms
内存限制
65536 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:
10 1 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4
N个结点，无序；
要求获得 左<中<=右的完全二叉树
首先输入sort排序非降#include<algorithm>
接着buildCBTreeDFS(CompleteBinarySearchTree*CBT, int STar ,int END_1)构建好完全二叉树，当前数组序号从STar到END_1；这里有len=END_1-STar+1个元素，
难点一int GETrightCount( int len)根据长度len获得这些子结点  组建成  子完全二叉树 右边的结点个数  并返回；
然后通过END_1-GETrightCount(len)获得root；
难点二：递归调用停止判断；
接着利用queue 先进先出#include<queue>  输出

评测结果
时间结果得分题目语言用时(ms)内存(kB)用户8月03日 00:13答案正确301064C++ (g++ 4.7.2)2436datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确138418/181答案正确14363/32答案正确13082/23答案正确13082/24答案正确13043/35答案正确23842/2

#include<iostream>   #include<algorithm>#include<queue>using namespace std;#define NOTNEXTNODEs -1struct CompleteBinarySearchTree{  int value;  int left_index;  int right_index;};int Log2_myself(int n){  int index = 0;  while (n > 0)  {    n=n >> 1;    index++;  }  return index;  /*n=n >> 1相当于n>>=1相当于n/2;  这里整个函数相当于求log2(n)；也可以用下面的  #include<math.h> (int)(log(x*1.0) / log(2.0))+1;*/}int pow_myself(int index,int radix){  int i,temp=index;  for (i = 1; i < radix; i++)    temp*= index;  return temp;/*#include<math.h>pow(index, radix)         等于index的radix次方*/}void Dureadln(CompleteBinarySearchTree*CBT,int N){  for (int index = 0; index < N; index++)  {    cin >> CBT[index].value;    CBT[index].left_index = CBT[index].right_index = NOTNEXTNODEs;  }}bool CBTCMp(const CompleteBinarySearchTree&A, const CompleteBinarySearchTree&B){  return A.value < B.value;}int GETrightCount( int len){   int level,lcount,lastfull;  level = Log2_myself(len);   lcount = pow_myself(2, level);   lastfull = lcount / 2;   if (lastfull/2 < len - lcount/2+1)     return  len - lcount / 2;   return lastfull / 2-1;}int buildCBTreeDFS(CompleteBinarySearchTree*CBT, int STar ,int END_1){  int root =END_1- GETrightCount(END_1-STar+1);     CBT[root].left_index = root >STar ? buildCBTreeDFS(CBT, STar, root-1) : NOTNEXTNODEs;   CBT[root].right_index = root < END_1 ? buildCBTreeDFS(CBT, root + 1, END_1) : NOTNEXTNODEs;   return root; }void printfCBT(int root, CompleteBinarySearchTree*CBT){  queue<int>index;  int size,newroot;  index.push(root);  cout << CBT[root].value;  while (!index.empty())  {    size = index.size();    while (size--)    {      root = index.front();      index.pop();      if (CBT[root].left_index != NOTNEXTNODEs)      {        newroot = CBT[root].left_index;        cout <<" "<< CBT[newroot].value;        index.push(newroot);      }      if (CBT[root].right_index!= NOTNEXTNODEs)      {        newroot = CBT[root].right_index;        cout << " " << CBT[newroot].value;        index.push(newroot);      }    }  }  cout << endl;}int main(){  CompleteBinarySearchTree*CBT;  int N;  cin >> N;  CBT= new CompleteBinarySearchTree[N];   Dureadln(CBT, N);  sort(CBT, CBT + N,CBTCMp);   printfCBT(buildCBTreeDFS(CBT, 0,N-1),CBT);  delete[]CBT;  system("pause");  return 0;}