PAT甲级——A1064 Complete Binary Search Tree

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

通过观察可以注意到，对完全二叉树当中的任何一个结点（设编号为x），其左孩子的编号一定是2x，而右孩子的编号一定是2x + 1。也就是说，完全二叉树可以通过建立一个大小为2k的数组来存放所有结点的信息，其中k为完全二叉树的最大高度，且1号位存放的必须是根结点（想一想为什么根结点不能存在下标为0处？）。这样就可以用数组的下标来表图95完全二又树编号示意示结点编号，且左孩子和右孩子的编号都可以直接计算得到。
事实上，如果不是完全二叉树，也可以视其为完全二叉树，即把空结点也进行实际的编号工作。但是这样做会使整棵树是一条链时的空间消耗巨大（对k个结点就需要大小为2^k的数组），因此很少采用这种方法来存放一般性质的树。不过如果题目中已经规定是完全二叉树，那么数组大小只需要设为结点上限个数加1即可，这将会大大节省编码复杂度。
除此之外，该数组中元素存放的顺序恰好为该完全二叉树的层序遍历序列。而判断某个结点是否为叶结点的标志为：该结点（记下标为root）的左子结点的编号root * 2大于结点总个数n（想一想为什么不需要判断右子结点？）；判断某个结点是否为空结点的标志为：该结点下标 root大于结点总个数n。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int N, nums[1001], res[1001], index = 0;
void levelOrder(int k)
{
    if (k > N)//叶子节点
        return;
    levelOrder(k * 2);//遍历左子树
    res[k] = nums[index++];//即遍历完左子树后，此时即为根节点
    levelOrder(k * 2 + 1);//遍历右子树
}
int main()
{
    cin >> N;
    for (int i = 0; i < N; ++i)
        cin >> nums[i];
    sort(nums, nums + N, [](int a, int b) {return a < b; });
    levelOrder(1);
    for (int i = 1; i <= N; ++i)
        cout << res[i] << (i == N ? "" : " ");
    return 0;
}