"""
File: space_complexity.py
Created Time: 2022-11-25
Author: Krahets (krahets@163.com)
"""

import sys, os.path as osp
sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
from modules import *

def function() -> int:
    """ 函数 """
    # do something
    return 0

def constant(n: int) -> None:
    """ 常数阶 """
    # 常量、变量、对象占用 O(1) 空间
    a: int = 0
    nums: List[int] = [0] * 10000
    node = ListNode(0)
    # 循环中的变量占用 O(1) 空间
    for _ in range(n):
        c: int = 0
    # 循环中的函数占用 O(1) 空间
    for _ in range(n):
        function()

def linear(n: int) -> None:
    """ 线性阶 """
    # 长度为 n 的列表占用 O(n) 空间
    nums: List[int] = [0] * n
    # 长度为 n 的哈希表占用 O(n) 空间
    mapp: Dict = {}
    for i in range(n):
        mapp[i] = str(i)

def linear_recur(n: int) -> None:
    """ 线性阶（递归实现） """
    print("递归 n =", n)
    if n == 1: return
    linear_recur(n - 1)

def quadratic(n: int) -> None:
    """ 平方阶 """
    # 二维列表占用 O(n^2) 空间
    num_matrix: List[List[int]] = [[0] * n for _ in range(n)]

def quadratic_recur(n: int) -> int:
    """ 平方阶（递归实现） """
    if n <= 0: return 0
    # 数组 nums 长度为 n, n-1, ..., 2, 1
    nums: List[int] = [0] * n
    return quadratic_recur(n - 1)

def build_tree(n: int) -> Optional[TreeNode]:
    """ 指数阶（建立满二叉树） """
    if n == 0: return None
    root = TreeNode(0)
    root.left = build_tree(n - 1)
    root.right = build_tree(n - 1)
    return root


""" Driver Code """
if __name__ == "__main__":
    n = 5
    # 常数阶
    constant(n)
    # 线性阶
    linear(n)
    linear_recur(n)
    # 平方阶
    quadratic(n)
    quadratic_recur(n)
    # 指数阶
    root: Optional[TreeNode] = build_tree(n)
    print_tree(root)