04-樹8. Complete Binary Search Tree (30)

時間限制

100 ms

內存限制

65536 kB

代碼長度限制

8000 B

判題程序

Standard

作者

CHEN, Yue

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

解題思路(轉載)：

這個題目要求構造完全二叉排序樹，如果直接採用普通的建樹方式，其實挺麻煩的，我剛開始也是那樣做得到，還調試了很久，AC了，但今天突然在網上看到別人做的這個題思路非常的好，所以消化吸收後寫了下，其實就是利用了數據結構中完全二叉樹的的一個性質：孩子節點的下標爲i則其左孩子節點的下標爲2*i，右孩子節點的下標爲2*i+1，這個性質只有完全二叉樹才滿足，其實反過來按這個性質構造出來的樹就是一個完全二叉樹。 要實現這棵完全二叉樹也是排序樹，其實就簡單了，顯然一棵二叉排序樹的中序遍歷序列是遞增有序的，所以這就簡單了，只要在構造完全二叉樹的時候按中序構造就可以了（前提是元素遞增有序）。

以前做樹的題目都是鏈接型的，而這個題目用完全二叉樹的這個性質確非常的合適，所以記錄下，供以後回顧學習。

<span style="font-size:18px;">#include <iostream>
#include <algorithm>
#define MAX 1001
using namespace std;

int node[MAX];
int tree[MAX];
int pos,n;

void InOrderBuild(int root)
{
    if(root>n)
        return;
    int lchild = root<<1;
    int rchild = (root<<1)+1;
    InOrderBuild(lchild);
    tree[root] = node[pos++];
    InOrderBuild(rchild);
}

int main()
{
    int i;
    cin>>n;
    for(i=0;i<n;i++)
        cin>>node[i];
    sort(node,node+n);
    pos = 0;
    InOrderBuild(1);
    for(i=1;i<n;i++)
        cout<<tree[i]<<" ";
    cout<<tree[i]<<endl;
    return 0;
}</span><span style="font-size:32px;color: rgb(255, 0, 0);">
</span>