PAT 1064. Complete Binary Search Tree (30)
2015-08-22 11:53
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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
#include <iostream> #include <vector> #include <algorithm> using namespace std; int n; vector<int> preOrder; void PreTraversal(int root) { if (root > n) return; PreTraversal(2 * root); preOrder.push_back(root); PreTraversal(2 * root + 1); } int main() { cin >> n; int num[1001], levelOrder[1001]; for (int i = 0; i < n; i++) cin >> num[i]; sort(num, num + n); PreTraversal(1); for (int i = 0; i < preOrder.size(); i++) levelOrder[preOrder[i]] = num[i]; for (int i = 1; i < n; i++) cout << levelOrder[i] << " "; cout << levelOrder ; }
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。
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