1064. Complete Binary Search Tree

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.

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

创建二叉搜索树，且该树同时为完全二叉树。

思路

创建BST一般方法是链表插入，但注意到是完全二叉树，其特殊性质是在数组中，若根节点下标为i，则2i为左子树根节点下标，2i+1为右子树根结点下标（两下标皆不超过节点数n），所以可以考虑使用数组进行创建比较简便。

要想创建树，要么知道中序序列以及其他遍历序列中的一种，要么知道一种遍历序列以及树的整体框架，显然本题是后面一种。由于二叉查找树的中序遍历序列是有序的，且完全二叉树的空框架可以用数组表示，那么只要在中序遍历时依次填入数据，即可构建完全二叉查找树。

代码实现

#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;

const int maxn = 1001;
int cbt[maxn], n;       // 完全二叉树框架，结点个数
int data[maxn], index = 0;      // 结点数据，已使用结点个数

void inOrder(int i)     // 中序遍历
{
    if (i > n)
        return;

    inOrder(2 * i);
    cbt[i] = data[index++];     // 访问根结点时填入数据
    inOrder(2 * i + 1);
}

void levelOrder()       // 层序遍历
{
    bool flag = true;       // flag用来控制空格输出
    queue<int> q;
    q.push(1);
    while(!q.empty())
    {
        int now = q.front();
        q.pop();
        if (flag)
            flag = false;
        else
            printf(" ");
        printf("%d", cbt[now]);
        if (2 * now <= n)
            q.push(2 * now);
        if (2 * now + 1 <= n)
            q.push(2 * now + 1);
    }
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d", &data[i]);
    sort(data, data + n);       // 读取完数据后排序，以便在中序遍历中使用

    inOrder(1);
    levelOrder();

    return 0;
}