I have a random binary tree in the following forms
12
13, 14
29, 26, 89
Each node has two children ie ( 12->(13, 14), 13->(29, 26), 14 ->(26, 89)). Here I need to return all possible paths in the forms [ [12, 13, 29], [ 12, 13, 26], [12, 14, 26], [12, 14, 89]]. I tried with the following code. I have problem with updating list. Thanks in advance.
class Tree:
def __init__(self, data, left=None, right=None):
self.data = data
self.left = left
self.right = right
def __str_(self):
return '%s' % self.data
def makeList(tree, path =[]):
if(tree != None):
path.append(tree.data)
if tree.left:
path.append(makeList(tree.left, path))
if tree.right:
path.append(makeList(tree.left, path))
return path
root = Tree(12)
root.left = Tree(13)
root.right = Tree(14)
root.right.left = Tree(26)
root.left.right = Tree(26)
root.left.left = Tree(29)
root.right.right = Tree(86)
x = makeList(root)
I don't know how to solve it using a memoized recursion . But I still post my answer since it may solve your problem partly.
def makeList(tree):
paths = []
if not (tree.left or tree.right):
return [[tree.data]]
if tree.left:
paths.extend([[tree.data] + child for child in makeList(tree.left)])
if tree.right:
paths.extend([[tree.data] + child for child in makeList(tree.right)])
return paths
When you set this:
def makeList(tree, path =[]):
This empty list after path is not deleted after the execution of the function, if you call makeList a second time without a path parameter, the path list you had at the end of the first call will remain.
Consider the function:
def f(l=[]):
l.append(17)
print l
If you keep on calling f() without the parameter l you will get another 17 each time.
[17]
[17, 17]
[17, 17, 17]
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