PAT-甲级-1064 Complete Binary Search Tree (30 分)

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本文介绍了一种解决完全二叉搜索树层序遍历问题的方法，利用BST的性质进行排序，并通过DFS遍历及CBT的节点编号规律，实现了从给定序列到层序遍历结果的有效转换。

1064 Complete Binary Search Tree (30 分)

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

题目大意：已知树是一棵完全二叉搜索树：即如果该层不是最后一层，则无叶子节点；最后一层没有只有右叶子无左叶子的情况。给定一个序列，输出该树的层序。

解题思路：

BST的性质：中序一定是有序序列->sort即可；
CBT的性质：节点编号满足父节点n，左孩子2*n，右孩子2*n+1（此时n=1）的规律；
用dfs遍历中序序列，再按照CBT的下标保存就可以得到层序。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define rep(i,j,k)    for(int i=j;i<k;i++)
const int maxn=1100;
int in[maxn],level[maxn];
int n,index=0;
void dfs(int u){
    if(u>n)    return;
    dfs(2*u);
    level[u] = in[index++];
    dfs(2*u+1);
}
int main(){
    std::ios::sync_with_stdio(false);
    cin>>n;
    rep(i,0,n)
    cin>>in[i];
    sort(in,in+n);
    dfs(1);
    rep(i,1,n+1)
    printf("%d%s",level[i],i==n?"":" ");
    return 0;
}