PAT甲级1064 log()函数的换底公式 BST 递归

题目

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

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题解

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
vector<int> in, level;
void levelorder(int start, int end, int index) {
    if(start > end) return ;
    int n = end - start + 1;
    int l =log(n+1)/log(2); // 如果不是满二叉树，这就是除了最后一层的层数,如果是满二叉树，就是二叉树层数。这里是log的换底公式
    int leave = n - (pow(2, l) - 1);// 最后一层的叶子节点数
    int root = start + (pow(2, l - 1) - 1) + min((int)pow(2, l - 1), leave); // pow(2, l - 1) - 1是除了root结点所在层和最后一层外，左子树的结点个数，pow(2, l - 1) 是根节点的左子树中，你要找的结点和最后一层夹着的几个结点，min(pow(2, l - 1), leave)是最后一层在根节点左子树的结点数
    level[index] = in[root];
    levelorder(start, root - 1, 2 * index + 1);
    levelorder(root + 1, end, 2 * index + 2);
}
int main() {
    int n;
    scanf("%d", &n);
    in.resize(n);
    level.resize(n);
    for(int i = 0 ; i < n; i++)
        scanf("%d", &in[i]);
    sort(in.begin(), in.end());
    levelorder(0, n - 1, 0);
    printf("%d", level[0]);
    for(int i = 1; i < n; i++)
        printf(" %d", level[i]);
    return 0;
}

给你一列数字比如0~9，告诉你这些组成了二叉搜索树，你怎么确定哪个是根结点。根本来说是这样一个问题。
我们可以画出这个二叉搜索树，它是一个完全二叉树，根节点的左子树也是一个完全二叉树，可以这样递归。