PAT 1064. Complete Binary Search Tree (30)

1064. Complete Binary Search Tree (30)

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: 
10
 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 int n;
 8 vector<int> preOrder;
 9 
10 void PreTraversal(int root)
11 {
12     if (root > n)
13         return;
14     PreTraversal(2 * root);
15     preOrder.push_back(root);
16     PreTraversal(2 * root + 1);
17 }
18 
19 int main()
20 {
21     cin >> n;
22 
23     int num[1001], levelOrder[1001];
24     for (int i = 0; i < n; i++)
25         cin >> num[i];
26 
27     sort(num, num + n);
28     PreTraversal(1);
29     for (int i = 0; i < preOrder.size(); i++)
30         levelOrder[preOrder[i]] = num[i];
31     for (int i = 1; i < n; i++)
32         cout << levelOrder[i] << " ";
33     cout << levelOrder[n];
34         
35 }

 

1 2 3 4 5 6 7 8 9 0
Sample Output: 
6 3 8 1 5 7 9 0 2 4

要求树为完全二叉树，可以参考二叉堆的方法，使用一个数组表示，根节点为1，然后对于根节点i的子节点为2*i，2*i+1。我们根据所获得的n构造一个level order为1, 2, 3, 4, ..., n-1, n的一颗完全二叉树，然后对其进行前序遍历，并对所有的数字进行从小到大排序，那么这样就可以得到每个数字在level order的索引。

posted @ 2015-08-22 11:53  JackWang822  阅读(666)  评论(0)  编辑  收藏  举报