🧠 Unlock the Slot Counting Trick for Valid Preorder Serialization
IntuitionIn a valid binary tree, every non-null node provides 2 new slots and takes 1. Null nodes (#) only take 1 slot. So we simulate this slot system while sc

### Intuition
In a valid binary tree, every non-null node provides 2 new slots and takes 1. Null nodes (`#`) only take 1 slot. So we simulate this slot system while scanning the input.

### Approach
![image.png](https://assets.leetcode.com/users/images/4769cb17-bebe-421f-9403-bab16effdcd9_1744265731.3607557.png)

Start with 1 available slot (for the root). As we process each node:
- Decrease slot count by 1 (node takes a slot).
- If it’s a non-null node, increase slot count by 2 (it adds two children).
- If at any point we have negative slots, it’s invalid.
- In the end, all slots should be used up exactly.

### Complexity
Time Complexity:  
( O(n) ) — iterate through all nodes once.

Space Complexity:  
( O(1) ) — just an integer counter for slots.

### Code
```python []
class Solution(object):
    def isValidSerialization(self, preorder):
        slots = 1
        for node in preorder.split(','):
            slots -= 1
            if slots < 0:
                return False
            if node != '#':
                slots += 2
        return slots == 0
```

```javascript []
function isValidSerialization(preorder) {
    let slots = 1;
    for (const node of preorder.split(",")) {
        slots--;
        if (slots < 0) return false;
        if (node !== "#") slots += 2;
    }
    return slots === 0;
}
```
<img src="https://assets.leetcode.com/users/images/b91de711-43d7-4838-aabe-07852c019a4d_1743313174.0180438.webp" width="270"/>
['String', 'Stack', 'Tree', 'Binary Tree', 'Python', 'JavaScript']