1064. Complete Binary Search Tree (30)-PAT甲级真题（树的层序遍历）

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：给一串构成树的序列，已知该树是完全二叉搜索树，求它的层序遍历的序列
分析：总得概括来说，已知中序，从根节点开始中序遍历，按中序数组给出的顺序依次将值填入level数组对应的下标中，输出level数组可得层序遍历。
1. 因为二叉搜索树的中序满足：是一组序列的从小到大排列，所以只需将所给序列排序即可得到中序数组in
2. 假设把树按从左到右、从上到下的顺序依次编号，根节点为0，则从根结点root = 0开始中序遍历，root结点的左孩子下标是root*2+1，右孩子下标是root*2+2
3. 因为是中序遍历，所以遍历结果与中序数组in中的值从0开始依次递增的结果相同，即in[t++]（t从0开始），将in[t++]赋值给level[root]数组
4. 因为树是按从左到右、从上到下的顺序依次编号的，所以level数组从0到n-1的值即所求的层序遍历的值，输出level数组即可～

#include <iostream>
#include <algorithm>
using namespace std;
int in[1010], level[1010], n, t = 0;
void inOrder(int root) {
    if (root >= n) return ;
    inOrder(root * 2 + 1);
    level[root] = in[t++];
    inOrder(root * 2 + 2);
}
int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d", &in[i]);
    sort(in, in + n);
    inOrder(0);
    printf("%d", level[0]);
    for (int i = 1; i < n; i++)
        printf(" %d", level[i]);
    return 0;
}

#include <iostream>

#include <algorithm>

using namespace std;

int in[1010], level[1010], n, t = 0;

void inOrder(int root) {

if (root >= n) return ;

inOrder(root * 2 + 1);

level[root] = in[t++];

inOrder(root * 2 + 2);

}

int main() {

scanf("%d", &n);

for (int i = 0; i < n; i++)

scanf("%d", &in[i]);

sort(in, in + n);

inOrder(0);

printf("%d", level[0]);

for (int i = 1; i < n; i++)

printf(" %d", level[i]);

return 0;

}

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