Complete Binary Search Tree

標籤：

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 (<=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

 解題思路：根據完全二叉樹性質找規律

 解題感悟：1、記住vector中的sort和cmath檔案中的一些函數很重要

               2、注意特殊情況下輸出的格式（本題的特殊情況是只有一個輸入的情況）

 1 #include <iostream> 2 #include <vector> 3 #include<algorithm> 4 #include <cmath>  5 using namespace std; 6 int main(){ 7     int n; 8     cin >> n; 9     vector<int> v;10     for (int i = 0; i < n; i++)11     {12         int number;13         cin >> number;14         v.push_back(number);15     }16     sort(v.begin(), v.end());17     int zhishu = log(n)/log(2);18     int tichushu = n -pow(2,zhishu)+ 1;19     vector<int> v1;20     for (int i = 0; i < tichushu;i++)21         v1.push_back(v[i * 2]);22     for (int i = 0; i < tichushu; i++)23         v.erase(v.begin() + i);24     int flag = 1;25     for (int i = 1; i <= zhishu;i++)26     {27         for (int j = 0; j < pow(2,i-1);j++){28             if (flag == 1)29             {30                 cout << v[v.size() / pow(2, i) + j * v.size() / pow(2, i-1)];31                 flag = 0;32             }33             else34             {35                 cout << " " << v[v.size() / pow(2, i) + j * v.size() / pow(2, i-1)];36             }    37         }38     }39     for (int i = 0; i < v1.size();i++)40     {41         if (flag == 1)42         {43             cout<< v1[i];44             flag = 0;45         }46         else47         cout << " " << v1[i];48     }49     return 0;50 }

 

Complete Binary Search Tree