題目

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

　　題目大意： 題目要求對給定的序列建立完全二叉搜尋樹，所謂完全二叉搜尋樹就是要滿足完全二叉樹的搜尋樹。 。
　　解題思路： 我們知道，完全二叉樹的結點i如果從1開始編號，那麼左兒子為2* i，右兒子為2*i+1；而二叉搜尋樹的中序遍歷為升序，因此只需要對輸入序列按照升序排序，然後對完全二叉樹進行中序遍歷，填入相應的元素即可。。

#include <iostream>
#include <vector>
#include <stdio.h>
#include <algorithm>
 
using namespace std;
 
vector<int> tree;
vector<int> nodes;
int N;
 
void buildTree(int root){
    static int index = 1;
    if(root > N) return;
    buildTree(root * 2);
    tree[root] = nodes[index++];
    buildTree(root * 2 + 1);
}
 
int main()
{
    cin >> N;
    nodes.resize(N+1);
    tree.resize(N+1);
    for(int i = 1; i <= N; i++){
        scanf("%d",&nodes[i]);
    }
    sort(nodes.begin(),nodes.end());
    buildTree(1);
    printf("%d",tree[1]);
    for(int i = 2; i <= N; i++)
        printf(" %d",tree[i]);
    cout << endl;
    return 0;
}