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 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4
题意
给出一个序列,要求输出这棵树的层序遍历,这棵树是完全二叉搜索树。
思路
二叉搜索树的中序遍历是有序的,所以只需要将这个序列排序就可以获得中序遍历。又因为这棵树是完全二叉树,所以每个节点的序号很容易确定。层序遍历实际上就是按照节点的序号排序。
代码
#include <algorithm>
#include <cstdio>
using namespace std;
const int MAX_SIZE = 1010;
int n, p = 0, nums[MAX_SIZE], cbt[MAX_SIZE];
void inOrder(int root) {
if (root > n)
return;
inOrder(root * 2);
cbt[root] = nums[p++];
inOrder(root * 2 + 1);
}
int main() {
scanf("%d", &n);
for (int i = 0; i < n; ++i)
scanf("%d", nums + i);
sort(nums, nums + n);
inOrder(1);
for (int i = 1; i <= n; ++i) {
printf("%d", cbt[i]);
if (i < n)
printf(" ");
}
}