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# Verify Preorder Serialization of a Binary Tree LeetCode Solution

## Problem – Verify Preorder Serialization of a Binary Tree LeetCode Solution

One way to serialize a binary tree is to use preorder 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 `'#'`. 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 `preorder`, return `true` if it is a correct preorder traversal serialization of a binary tree.

It is guaranteed that 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"`.

Note: You are not allowed to reconstruct the tree.

Example 1:

``````Input: preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true
``````

Example 2:

``````Input: preorder = "1,#"
Output: false
``````

Example 3:

``````Input: preorder = "9,#,#,1"
Output: false
``````

Constraints:

• `1 <= preorder.length <= 104`
• `preorder` consist of integers in the range `[0, 100]` and `'#'` separated by commas `','`.

## Verify Preorder Serialization of a Binary Tree LeetCode Solution in Java

``````public boolean isValidSerialization(String preorder) {
String[] nodes = preorder.split(",");
int diff = 1;
for (String node: nodes) {
if (--diff < 0) return false;
if (!node.equals("#")) diff += 2;
}
return diff == 0;
}
``````

## Verify Preorder Serialization of a Binary Tree LeetCode Solution in Python

``````class Solution(object):
def isValidSerialization(self, preorder):
"""
:type preorder: str
:rtype: bool
"""
# remember how many empty slots we have
# non-null nodes occupy one slot but create two new slots
# null nodes occupy one slot

p = preorder.split(',')

#initially we have one empty slot to put the root in it
slot = 1
for node in p:

# no empty slot to put the current node
if slot == 0:
return False

# a null node?
if node == '#':
# ocuppy slot
slot -= 1
else:
# create new slot
slot += 1

#we don't allow empty slots at the end
return slot==0
``````

## Verify Preorder Serialization of a Binary Tree LeetCode Solution in C++

``````class Solution {
public:
bool isValidSerialization(string preorder) {
if (preorder.empty()) return false;
preorder+=',';
int sz=preorder.size(),idx=0;
int capacity=1;
for (idx=0;idx<sz;idx++){
if (preorder[idx]!=',') continue;
capacity--;
if (capacity<0) return false;
if (preorder[idx-1]!='#') capacity+=2;
}
return capacity==0;
}
};
``````
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