PAT A1064 Complete Binary Search Tree GO语言题解及注意事项

1064 Complete Binary Search Tree (30分)

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

 

思路:

　　由于是完全二叉树，所以可以用大小为n+1的数组保存，其下标为[1,n]，Tree[i]的左子树是Tree[2*i]（右子树是Tree[2*i+1]），当2*i（或2*i+1）>n时，说明Tree[i]没有左子树（右子树）此时数组的顺序遍历就是完全二叉树的层序遍历，并且我们可以得到该树的形状：

 

　　而该树又是一颗二叉搜索数，那么他的中序遍历就是一个递增数列，由此得出，只要将给定序列由小到大排序，就得到了该树的中序遍历序列，中序遍历该树并将数的中序遍历序列插入对应节点，即可得建树；

 

Solution(Golang):

package main

import (
	"fmt"
	"sort"
)

var (
	bts  = make([]int, 10010)
	cnt  = 0
	n    int
	nums = make([]int, 10010)
)

func inorder(index int) {
	if index > n {
		return
	}
	inorder(index * 2)
	bts[index] = nums[cnt]
	cnt++
	inorder(index*2 + 1)
}

func main() {
	fmt.Scan(&n)
	for i := 0; i < n; i++ {
		fmt.Scan(&nums[i])
	}
	sort.Ints(nums[0:n])
	inorder(1)
	for i := 1; i <= n; i++ {
		fmt.Print(bts[i])
		if i < n {
			fmt.Print(" ")
		}
	}
}