2b5bbb98ca
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.
49 lines
935 B
Python
49 lines
935 B
Python
from collections import deque
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class Node(object):
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def __init__(self, value, left=None, right=None):
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self.value = value
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self.left = left
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self.right = right
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def is_bst(self):
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s = []
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s.append((float('-inf'), self, float('inf')))
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while s:
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lo, node, hi = s.pop()
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if lo <= node.value <= hi:
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node.left and s.append((lo, node.left, node.value))
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node.right and s.append((node.value, node.right, hi))
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else:
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return False
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return True
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x = Node(
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50,
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Node(
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17,
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Node(
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12,
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Node(9),
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Node(14),
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),
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Node(
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23,
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Node(19),
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),
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),
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Node(
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72,
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Node(
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54,
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None,
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Node(67)
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),
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Node(76),
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),
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)
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assert x.is_bst()
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print("Success!")
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