04-树6 Complete Binary Search Tree

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:

10 1 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

解:这是一个完全二叉树问题，我在自己独立思考的情况下，想不到合适的方法，在看了解法之后，我震惊了：按照题意，将输入的数组由小到大排序之后就是二叉树的中序遍历情况，现在想要知道逐层遍历的情况，只需要将二叉树的下标进行中序遍历，对应即可解出每一个下标在中序遍历数组中对应的数。看代码：

#include <iostream> #include <stdlib.h> using namespace std; const int maxlen = 10001; int in_index = 0; int level[maxlen]; int compare(const void *a, const void *b){ return *(int*)a - *(int*)b; } void in_pass(int root, int N, int array[]){ if(root <= N){ in_pass(root*2, N, array); level[root] = array[in_index++]; in_pass(root*2+1, N, array); } } int main(){ int N; cin>>N; int in[maxlen]; for(int i = 0; i < N; i++){ cin>>in[i]; } qsort(in, N, sizeof(int), compare); // for(int i = 0; i < N; i++){ // cout<<in[i]<<' '; // } // cout<<endl; in_pass(1, N, in); cout<<level[1]; for(int i = 2; i <= N; i++){ cout<<' '<<level[i]; } return 0; }