04-树6 Complete Binary Search Tree

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

题目思路：

要求构造完全二叉排序树，如果直接采用普通的建树方式，挺麻烦，可以

利用了数据结构中完全二叉树的的一个性质：孩子节点的下标为i则其左孩子节点的下标为2*i，右孩子节点的下标为2*i+1，这个性质只有完全二叉树才满足。

#include<stdio.h>#include<stdlib.h>#define Naxsize 1000void solve(int Aleft, int Aright,int TRoot, int A[],int T[]);int compare(const void *a, const void *b);  int Get_Left_Nodes(int n);  int Min(int a, int b);  int main(){//freopen("test.txt", "r", stdin); //一共多少个节点int N;scanf("%d\n",&N);//接收收入的节点int A[N],T[N];for(int i = 0; i < N; i++){scanf("%d",&A[i]);} //调整为完全二插搜索树,//排序成中序遍历 qsort(A,N,sizeof(int),compare);int ALeft = 0, ARight = N -1, TRoot = 0;      solve(ALeft, ARight, TRoot, A, T); //输出for (int i = 0; i < N; ++i){          if(i == 0){              printf("%d", T[i]);          }          else              printf(" %d", T[i]);      }    return 0;} int compare(const void *a, const void *b)  {      return *(int*)a - *(int*)b;  }      void solve(int Aleft, int Aright, int TRoot,int A[],int T[]){/*初始调用为*solve(0,N-1,0)*/ int n;//看这里面还有元素嘛 n = Aright - Aleft+1;if(n == 0){return;} //计算出n个节点的树其左子树有多少个节点 int L,LeftRoot,RightRoot;L = Get_Left_Nodes(n);T[TRoot] = A[Aleft + L];//根节点的值 //在堆里面，开始的下标是1，0不存在真实值，所以左儿子是2i，这里是2i+1 LeftRoot = TRoot * 2 + 1;RightRoot = LeftRoot + 1; solve(Aleft,Aleft+L-1,LeftRoot,A,T);solve(Aleft+L+1,Aright,RightRoot,A,T);}int Get_Left_Nodes(int n)  {      int H = 0, tmp = 1, X, L;//X为左子树最下一层的结点数       int N = n;      while(N > 1){          N /= 2;          H++;//树的高度      }      for (int i = 0; i < H - 1; ++i){          tmp *= 2;      }      X = n - 2 * tmp + 1;      X = Min( X, tmp );      L = tmp - 1 + X;      return L;  }    int Min(int a, int b)  {      return (a < b) ? a : b;  }