PAT 甲级 1064 Complete Binary Search Tree

https://pintia.cn/problem-sets/994805342720868352/problems/994805407749357568

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

 

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N;
int a[maxn];
vector<int> level;

void levelorder(int st, int en, int index) {
    if(st > en) return ;
    int n = en - st + 1;
    int l = log(n + 1) / log(2);
    int leave = n - (pow(2, l) - 1);
    int root = st + (pow(2, l - 1) - 1) + min((int)pow(2, l - 1), leave);
    level[index] = a[root];
    levelorder(st, root - 1, 2 * index + 1);
    levelorder(root + 1, en, 2 * index + 2);
}

int main() {
    scanf("%d", &N);
    level.resize(N);
    for(int i = 0; i < N; i ++)
        scanf("%d", &a[i]);

    sort(a, a + N);
    levelorder(0, N - 1, 0);
    for(int i = 0; i < N; i ++) {
        printf("%d", level[i]);
        printf("%s", i != N - 1 ? " " : "
");
    }
    return 0;
}

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