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

思路：

1、首先因为是完全二叉树，因此可以用数组来存，既方便又不浪费空间，而且数组就是按层次遍历的顺序存的，最后只要把数组顺序输出就行了。

2、因为又是二叉搜索树，所以左边的结点一定小于根结点；再根据完全二叉树的性质，能够算出根节点是哪个元素

3、用分而治之的方法，先求出整棵树的根结点；然后再分别求左子树的根结点，再求右子树的根节点。

注意点：

如何根据二叉搜索树和完全二叉树的性质，来求出根结点

1、完全二叉树，知道了结点个数，就能确定整棵树的形状；因而能求出左子树有多少个结点

2、二叉搜索树，左边的结点一定小于根结点，又因为左边的结点个数L已经算出，只要将所给的元素从小到大排序，第L+1个就是根结点。

#include <stdlib.h>#include <stdio.h>#include <math.h>int arr[1005], rearr[1005];int compare( const void* a, const void* b ){return *(int*)a - *(int*)b;}int getLeftLength(int n){    //利用二叉树的性质：满二叉树第i层有 2^(i-1) 个结点， 高为h的满二叉树有 2^h - 1 个结点（从1开始）double h, x, L, t;h = (double)(int)( log((double)n+1) / log(2.0) );//h = floor( log((double)n+1) / log(2.0) );x = n - pow(2.0, h) + 1 ;t = pow(2.0, h - 1.0);x = x < t ? x : t;L = t - 1 + x;return (int)L;}void solve( int left, int right, int root ){//初始调用： solve(0, n-1, 0);int n, L, leftRoot, rightRoot;n = right - left + 1;//数组中的总个数if(n == 0) return ;//递归退出的条件L = getLeftLength(n);//计算出左子树的结点rearr[root] = arr[left + L];//将新的根结点放入新的数组leftRoot = root * 2 + 1;//左孩子rightRoot = leftRoot + 1;//右孩子solve(left, left + L - 1, leftRoot);solve(left + L + 1, right, rightRoot);}int main(){int n;scanf("%d", &n);for(int i = 0; i < n; i++){scanf("%d", &arr[i]);}qsort(arr, n, sizeof(int), compare);solve(0, n-1, 0);for(int i = 0; i < n; i++){if( i != 0 ) printf(" ");printf("%d", rearr[i]);}system("pause");return 0;}