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verify-preorder-serialization-of-a-binary-tree.py
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verify-preorder-serialization-of-a-binary-tree.py
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# Time: O(n)
# Space: O(1)
# One way to serialize a binary tree is to use pre-oder traversal.
# When we encounter a non-null node, we record the node's value.
# If it is a null node, we record using a sentinel value such as #.
#
# _9_
# / \
# 3 2
# / \ / \
# 4 1 # 6
# / \ / \ / \
# # # # # # #
# For example, the above binary tree can be serialized to the string
# "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.
#
# Given a string of comma separated values, verify whether it is a
# correct preorder traversal serialization of a binary tree.
# Find an algorithm without reconstructing the tree.
#
# Each comma separated value in the string must be either an integer
# or a character '#' representing null pointer.
#
# You may assume that the input format is always valid, for example
# it could never contain two consecutive commas such as "1,,3".
#
# Example 1:
# "9,3,4,#,#,1,#,#,2,#,6,#,#"
# Return true
#
# Example 2:
# "1,#"
# Return false
#
# Example 3:
# "9,#,#,1"
# Return false
class Solution(object):
def isValidSerialization(self, preorder):
"""
:type preorder: str
:rtype: bool
"""
def split_iter(s, tok):
start = 0
for i in xrange(len(s)):
if s[i] == tok:
yield s[start:i]
start = i + 1
yield s[start:]
if not preorder:
return False
depth, cnt = 0, preorder.count(',') + 1
for tok in split_iter(preorder, ','):
cnt -= 1
if tok == "#":
depth -= 1;
if depth < 0:
break
else:
depth += 1
return cnt == 0 and depth < 0