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This article is about how to verify the preorder serialization of the python binary tree, the editor thinks it is very practical, so I share it with you to learn. I hope you can get something after reading this article.

I. the content of the topic

One way to serialize a binary tree is to use preorder traversal. When we encounter a non-empty node, we can record the value of that node. If it is an empty node, we can use a tag value record, such as #.

_ 9_

/\

3 2

/ /

4 1 # 6

/ /

# #

For example, the above binary tree can be serialized into the string "9pr 3pj4", where # represents an empty node.

Given a comma-separated sequence, verify that it is the correct preorder serialization of the binary tree. Write a feasible algorithm without reconstructing the tree.

Each comma-separated character is either an integer or a'# 'that represents a null pointer.

You can think that the input format is always valid, for example, it will never contain two consecutive commas, such as "1thecomponent 3".

Example 1:

Enter: "9, 3, 4, 4, 1, 1, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Output: true

Example 2:

Enter: "1Jing #"

Output: false

Example 3:

Enter: "9 minutes, 9 minutes, 1 years."

Output: false

Second, the way to solve the problem

Use the stack to store numbers, and'#'as the basis for getting out of the stack

Store a'#'in the stack beforehand

When the last one in the stack is'# 'and the last preorder sequence is' #', it means that the preorder serialization of the binary tree is correct.

Otherwise, if the'#'in preorder matches the'#'in the stack before it reaches the last position, it will be incorrect.

Code class Solution: def isValidSerialization (self, preorder: str)-> bool: stack = ['#'] I = 0 while I < len (preorder): if preorder [I] = =' ': I + = 1 continue elif preorder [I] =' #': if stack [- 1] = ='#': if I = = len (preorder)-1: stack.pop () else: return False Else: stack.pop () elif preorder [I]! ='#': stack.append (preorder [I]) while I < len (preorder) and preorder [I]! =' ': I + = 1 I + = 1 return len (stack) = = 0if _ _ name__ = =' _ _ main__': s = Solution () preorder = "9, main__':, 3, 4, 2, 6, 6, 5, 6, 6, 5, 6, 6, 6, 6, 6, 6, 6, 6, 5, 6, 6, 6, 5, 6, 6, 5, 6, 5, 5, 6, 5, 5, 5, 5, 5, 6, 1, 6, 6, 5, 5, 6, 1, 5, 6, 1, 5, 6, 6, 1, 5, 6, 6, 1, 5, 6, 1, 5, 6, 6, 1, 5, 6, 1, 5, 6, 1, 5, 6, 1, 6, 6, 1, 2, 6, 6, 1, 5, 6, 1, 5, 5, The editor believes that there are some knowledge points that we may see or use in our daily work. I hope you can learn more from this article. For more details, please follow the industry information channel.

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