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
           

題意

建立二叉搜尋樹，且該樹同時為完全二叉樹。

思路

建立BST一般方法是連結清單插入，但注意到是完全二叉樹，其特殊性質是在數組中，若根節點下标為i，則2i為左子樹根節點下标，2i+1為右子樹根結點下标（兩下标皆不超過節點數n），是以可以考慮使用數組進行建立比較簡便。

要想建立樹，要麼知道中序序列以及其他周遊序列中的一種，要麼知道一種周遊序列以及樹的整體架構，顯然本題是後面一種。由于二叉查找樹的中序周遊序列是有序的，且完全二叉樹的空架構可以用數組表示，那麼隻要在中序周遊時依次填入資料，即可建構完全二叉查找樹。

代碼實作

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

const int maxn = 1001;
int cbt[maxn], n;       // 完全二叉樹架構，結點個數
int data[maxn], index = 0;      // 結點資料，已使用結點個數

void inOrder(int i)     // 中序周遊
{
    if (i > n)
        return;

    inOrder(2 * i);
    cbt[i] = data[index++];     // 通路根結點時填入資料
    inOrder(2 * i + 1);
}

void levelOrder()       // 層序周遊
{
    bool flag = true;       // flag用來控制空格輸出
    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);       // 讀取完資料後排序，以便在中序周遊中使用

    inOrder(1);
    levelOrder();

    return 0;
}