[英]Tree Traversals, recursion is faster than iteration in python?
I have implement tree preorder traversal in python, but found that my recursive version is faster than iteration version. 我在python中实现了树前序遍历,但发现我的递归版本比迭代版本更快。
code is as below: 代码如下:
from __future__ import print_function
import time
class Tree():
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def build_tree(string):
nodes = [0] + [Tree(s) for s in string]
for i in range(2, len(nodes)):
p = i/2
if i%2 == 0:
nodes[p].left = nodes[i]
else:
nodes[p].right = nodes[i]
return nodes[1]
def preorder(tree):
if tree:
# print(tree.value,end='')
preorder(tree.left)
preorder(tree.right)
def preorder2(tree):
t = tree
s = []
while t or s:
while t:
# print(t.value,end='')
s.append(t)
t = t.left
if s:
t = s.pop()
t = t.right
def main():
tree = build_tree('abcdefghi'*1000)
t = time.time()
for _ in range(100):
preorder(tree)
print(time.time() - t)
t = time.time()
for _ in range(100):
preorder2(tree)
print(time.time() - t)
if __name__ == '__main__':
main()
results: 结果:
0.751042842865
1.0220580101
It means recursive version is about 25% faster. 这意味着递归版本的速度提高了约25%。 I search a lot, everybody says recursive should be slower, I just wonder why it is not the case in my code? 我搜索了很多,每个人都说递归应该慢一点,我只是想知道为什么我的代码不是这样的?
I believe you can simplify the iterator function and reduce the timing by eliminating one of the variables. 我相信你可以通过消除其中一个变量来简化迭代器功能并减少时间。 Also, deque
performs better than a set
or a list
in those kinds of cases. 此外,在这种情况下, deque
性能优于set
或list
。
from collections import deque
def preorder3(initial_node):
queue = deque([initial_node])
while queue:
node = queue.pop()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
The benchmarks: 基准:
from __future__ import print_function
from timeit import timeit
class Tree():
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def build_tree(string):
nodes = [0] + [Tree(s) for s in string]
for i in range(2, len(nodes)):
p = i/2
if i%2 == 0:
nodes[p].left = nodes[i]
else:
nodes[p].right = nodes[i]
return nodes[1]
def preorder(tree):
if tree:
preorder(tree.left)
preorder(tree.right)
def preorder2(tree):
t = tree
s = []
while t or s:
while t:
s.append(t)
t = t.left
if s:
t = s.pop()
t = t.right
from collections import deque
def preorder3(initial_node):
queue = deque([initial_node])
while queue:
node = queue.pop()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
tree = build_tree('abcdefghi'*100)
# Repetitions to time
number = 20
# Time it
print('preorder: ', timeit('f(t)', 'from __main__ import preorder as f, tree as t', number=number))
print('preorder2: ', timeit('f(t)', 'from __main__ import preorder2 as f, tree as t', number=number))
print('preorder3: ', timeit('f(t)', 'from __main__ import preorder3 as f, tree as t', number=number))
Prints: 打印:
preorder: 0.0256490707397
preorder2: 0.0419111251831
preorder3: 0.0269520282745
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.