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day18: there must be a better way to find left and right numbers instead of traversing the full tree again and again ...

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Stefan Harmuth 2021-12-21 05:58:37 +01:00
parent 81bef8d2bf
commit 0ef34d1a32
1 changed files with 24 additions and 12 deletions
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30
day18.py
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@ -115,8 +115,21 @@ class BinarySnailfish:
return root
def getLeftNumber(self) -> Union[None, 'BinarySnailfish']:
parent = self.parent
if self == self.parent.left:
while parent.parent and parent == parent.parent.left:
parent = parent.parent
parent = parent.left
else:
while parent.parent and parent == parent.parent.right:
parent = parent.parent
parent = parent.right
while parent.right.value is None:
parent = parent.right
return parent
leftnum = None
for current in self.getRoot().traverse_inorder():
for current in self.getRoot().traverse_preorder():
if current == self:
return leftnum
if current.value is not None:
@ -124,7 +137,7 @@ class BinarySnailfish:
def getRightNumber(self) -> Union[None, 'BinarySnailfish']:
returnNext = 0
for current in self.getRoot().traverse_inorder():
for current in self.getRoot().traverse_preorder():
if returnNext >= 3 and current.value is not None:
return current
if returnNext > 0:
@ -150,14 +163,13 @@ class BinarySnailfish:
self.value = 0
def reduce(self):
print(self.fours)
if self.value is not None:
return
found = True
while found:
found = False
for n in self.fours:
self.fours.pop(0)
for n in self.traverse_preorder():
if n.value is None and n.depth == 4:
n.explode()
found = True
break
@ -165,21 +177,21 @@ class BinarySnailfish:
if found:
continue
for n in self.traverse_inorder():
for n in self.traverse_preorder():
if n.value is not None and n.value > 9:
n.split()
found = True
break
def traverse_inorder(self):
def traverse_preorder(self):
yield self
if self.left:
for x in self.left.traverse_inorder():
for x in self.left.traverse_preorder():
yield x
if self.right:
for x in self.right.traverse_inorder():
for x in self.right.traverse_preorder():
yield x
def __add__(self, other):