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The idea is to track available slots in the binary tree. Initially, we start with one slot for the root. For each node, we consume a slot. If the node is non-null, it adds two more slots (one for each child). A null node ('#') does not add any slots. The binary tree is valid if, at the end of processing all nodes, all slots are exactly filled (i.e., zero slots remaining).
Time Complexity: O(n), where n is the length of the preorder string, as we are iterating through the nodes once.
Space Complexity: O(1), with no additional data structures used.
1class Solution:
2 def isValidSerialization(self, preorder: str) -> bool:
3 slots = 1
4 nodes = preorder.split(',')
5 for node in nodes:
6 slots -= 1
7 if slots < 0:
8 return False
9 if node != '#':
10 slots += 2
11 return slots == 0
12
13# Example Usage
14solution = Solution()
15print(solution.isValidSerialization("9,3,4,#,#,1,#,#,2,#,6,#,#"))In this Python implementation, the preorder string is split into nodes. The slots management is handled as per the provided logic: decrementing for each node and incrementing if the node is not null.
This approach involves simulating a stack-based tree traversal. The idea is to treat each "open bracket" as requiring nodes and each "close bracket" as completing them. We aim to manage the state of openings and closures using integers, representing how many nodes are expected at any point in the traversal.
Time Complexity: O(n), parsing each node once.
Space Complexity: O(1), with only integer variables used for tracking.
The Python implementation maintains a shift in expected nodes count, incrementing and decrementing for non-null and null nodes, respectively.