04-树6 Complete Binary Search Tree
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A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
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:
101 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;}
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