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.
Each input file contains one test case. For each case, the first line contains a positive integer N (≤). 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.
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.
10
1 2 3 4 5 6 7 8 9 0
6 3 8 1 5 7 9 0 2 4
题目分析:刚开始我想的是先把末尾的多余的元素 计入计算根节点的位置 但欠缺考虑到对于k层树来说
若第k层元素大于2的k-1次 我这样的做法就出了问题
只过了2个点(21分) //给分真高
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <climits> 3 #include<iostream> 4 #include<vector> 5 #include<queue> 6 #include<map> 7 #include<set> 8 #include<stack> 9 #include<algorithm> 10 #include<string> 11 #include<cmath> 12 using namespace std; 13 int Array[1010]; 14 int Queue[1010]; 15 void EnQueue(int begin, int end,int pos) //左闭右开 16 { 17 if (begin >= end) 18 return; 19 Queue[pos] = Array[(begin + end) / 2]; 20 EnQueue(begin, (begin + end)/2, 2 * pos + 1); 21 EnQueue((begin + end) / 2 + 1, end, 2 * pos + 2); 22 } 23 int main() 24 { 25 int N; 26 cin >> N; 27 for (int i = 0; i < N; i++) 28 cin >> Array[i]; 29 sort(Array, Array + N); 30 int sum = 1; 31 for (; sum * 2 < N; sum *= 2); 32 int offset = (N - sum)/2; 33 int i = 0; 34 Queue[i] = Array[N / 2 + offset]; //根节点入队 35 //分别递归处理左右 36 EnQueue(0, N / 2 + offset, 2 * i + 1); 37 EnQueue(N / 2 + offset + 1,N,2 * i + 2); 38 for (int i = 0; i < N - 1; i++) 39 cout << Queue[i] << " "; 40 if(N!=0) 41 cout << Queue[N- 1]; 42 return 0; 43 }
参考别人的做法
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <climits> 3 #include<iostream> 4 #include<vector> 5 #include<queue> 6 #include<map> 7 #include<set> 8 #include<stack> 9 #include<algorithm> 10 #include<string> 11 #include<cmath> 12 using namespace std; 13 int Array[1010]; 14 int Queue[1010]; 15 int N; 16 int id; 17 void EnQueue(int root) //左闭右开 18 { 19 if (root >= N) 20 return; 21 EnQueue(2 * root + 1); 22 Queue[root] = Array[id++]; 23 EnQueue(2 * root + 2); 24 } 25 int main() 26 { 27 cin >> N; 28 for (int i = 0; i < N; i++) 29 cin >> Array[i]; 30 sort(Array, Array + N); 31 EnQueue(0); 32 for (int i = 0; i < N - 1; i++) 33 cout << Queue[i] << " "; 34 cout << Queue[N - 1]; 35 return 0; 36 }
来自https://blog.csdn.net/feng_zhiyu/article/details/82219702
果然 很多代码真的是又短又好 虽然原理简单 但体现出的思想正是让我感到震撼
1064 Complete Binary Search Tree (30分)（已知中序输出层序遍历)
原文：https://www.cnblogs.com/57one/p/12063555.html