1064 Complete Binary Search Tree (30分)（已知中序输出层序遍历)

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 (≤). 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
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

 #define _CRT_SECURE_NO_WARNINGS
 #include <climits>
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<algorithm>
 #include<string>
 #include<cmath>
 using namespace std;
 int Array[];
 int Queue[];
 void EnQueue(int begin, int end,int pos) //左闭右开
 {
     if (begin >= end)
         return;
     Queue[pos] = Array[(begin + end) / ];
     EnQueue(begin, (begin + end)/,  * pos + );
     EnQueue((begin + end) /  + , end,  * pos + );
 }
 int main()
 {
     int N;
     cin >> N;
     for (int i = ; i < N; i++)
         cin >> Array[i];
     sort(Array, Array + N);
     int sum = ;
     for (; sum *  < N; sum *= );
     int offset = (N - sum)/;
     int i = ;
     Queue[i] = Array[N /  + offset]; //根节点入队
     //分别递归处理左右
     EnQueue(, N /  + offset,  * i + );
     EnQueue(N /  + offset + ,N, * i + );
     for (int i = ; i < N - ; i++)
         cout << Queue[i] << " ";
     if(N!=)
         cout << Queue[N- ];
     return ;
 }

参考别人的做法

 #define _CRT_SECURE_NO_WARNINGS
 #include <climits>
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<algorithm>
 #include<string>
 #include<cmath>
 using namespace std;
 int Array[];
 int Queue[];
 int N;
 int id;
 void EnQueue(int root) //左闭右开
 {
     if (root >= N)
         return;
     EnQueue( * root + );
     Queue[root] = Array[id++];
     EnQueue( * root + );
 }
 int main()
 {
     cin >> N;
     for (int i = ; i < N; i++)
         cin >> Array[i];
     sort(Array, Array + N);
     EnQueue();
     for (int i = ; i < N - ; i++)
         cout << Queue[i] << " ";
     cout << Queue[N - ];
     return ;
 }

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