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Refactor existing bst-checker implementation

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I believe the previous solution is invalid. This solution works and it should be
more time and space efficient.

Space-wise our stack grows proportionate to the depth of our tree, which for a
"balanced" BST should be log(n). Doing a BFT on a BST results in memory usage of
n because when we encounter the leaf nodes at the final level in the tree, they
will be 1/2 * n for a balanced BST.
This commit is contained in:
William Carroll 2020-11-21 14:14:50 +00:00
parent 1dc6695a47
commit 2b5bbb98ca
1 changed files with 10 additions and 40 deletions
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50
scratch/facebook/bst-checker.py
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@ -6,47 +6,17 @@ class Node(object):
self.left = left self.left = left
self.right = right self.right = right
def insert_left(self, value):
self.left = Node(value)
return self.left
def insert_right(self, value):
self.right = Node(value)
return self.right
def min(self):
xs = deque()
result = float('inf')
xs.append(self)
while xs:
node = xs.popleft()
result = min(result, node.value)
if node.left:
xs.append(node.left)
if node.right:
xs.append(node.right)
return result
def max(self):
xs = deque()
result = float('-inf')
xs.append(self)
while xs:
node = xs.popleft()
result = max(result, node.value)
if node.left:
xs.append(node.left)
if node.right:
xs.append(node.right)
return result
def is_bst(self): def is_bst(self):
result = True s = []
if self.left: s.append((float('-inf'), self, float('inf')))
result = result and self.left.max() < self.value while s:
if self.right: lo, node, hi = s.pop()
result = result and self.right.min() > self.value if lo <= node.value <= hi:
return result node.left and s.append((lo, node.left, node.value))
node.right and s.append((node.value, node.right, hi))
else:
return False
return True
x = Node( x = Node(