Verify preorder serialization of a binary tree¶
Time: O(N); Space: O(1); medium
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:
Input: “9,3,4,#,#,1,#,#,2,#,6,#,#”
Output: True
Example 2:
Input: “1,#”
Return False
Example 3:
Input: “9,#,#,1”
Output: False
[1]:
class Solution1(object):
def isValidSerialization(self, preorder: str) -> bool:
"""
:type preorder: str
:rtype: bool
"""
def split_iter(s, tok):
start = 0
for i in range(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
[2]:
s = Solution1()
preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#"
assert s.isValidSerialization(preorder) == True
preorder = "1,#"
assert s.isValidSerialization(preorder) == False
preorder = "9,#,#,1"
assert s.isValidSerialization(preorder) == False