Leetcode: Verify Preorder Serialization of a Binary Tree

Verify Preorder Serialization of a Binary Tree



Similar Problems:


One way to serialize a binary tree is to use pre-order 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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/verify-preorder-serialization-of-a-binary-tree
## Basic Ideas: Stack
##      If number, push.
##      If #, check stack head. 
##         If empty, return false; 
##         If #, the next element should be a node value. If not, return False
##         Pop them both. Keep checking, then push #
##      At the end, stack should have only one element. And it's #
##
##  Sample data: "#"
## Complexity: Time O(n), Space O(n)
class Solution(object):
    def isValidSerialization(self, preorder):
        """
        :type preorder: str
        :rtype: bool
        """
        stack = []
        for element in preorder.split(','):
            if element != '#':
                stack.append(element)
            else:
                while True:
                    if len(stack) <= 1:
                        stack.append('#')
                        break
                    if stack[-1] != '#':
                        stack.append('#')
                        break
                    if stack[-1] == '#' and stack[-2] == '#': return False
                    # pop #, then pop value. Then plan to insert another '#'
                    stack.pop()
                    stack.pop()
                else:
                    stack.append(element)
        return stack == ['#']
linkedin
github
slack

Share It, If You Like It.

Leave a Reply

Your email address will not be published.