1064. Complete Binary Search Tree (30)

1064. Complete Binary Search Tree (30)

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

题意：

给定n个数，先建立完全排序二叉树（即节点的左孩子中的所有节点均小于该节点，节点的右孩子中的所有节点均大于该节点），再按照广度优先遍历的方式输出每个节点。

      刚开始写这道题的时候，希望模拟出来这道题，发现非常麻烦，首先要排序，再建立一棵完全排序二叉树，再用广搜依次输出每个节点。后来代码敲着敲着，发现，这样好像会超时，而且貌似有非常简便的方法。

      事实上、我们可以通过中序遍历直接求解这道题。因为完全排序二叉树的排序方法，它的结果输出就是通过中序排序实现的（满足左孩子<父节点<右孩子），故中序遍历的序列即为排好序的序列。

     因此，我们只需要将原来已经排好序的序列按照中序遍历存储，即为答案，同样，输出的结果再按照中序遍历的方式输出，即为排好序的序列。

代码：

#include 
#include 
#include 
using namespace std;
int n,a[1005],b[1005],index,num=0;
void dfs(int odex)
{
    if(odex>=n)
    {
        num++;
        return;
    }
    dfs(2*odex+1);
    b[odex]=a[num-1];
    dfs(2*odex+2);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0; i

结果：