Problem Description

You are given a string that represents a preorder traversal serialization of a binary tree. In this serialization:

• Each non-null node is represented by its integer value
• Each null node is represented by the character '#'
• Values are separated by commas

For example, a binary tree with root 9, left child 3, and right child 2 could be serialized as "9,3,4,#,#,1,#,#,2,#,6,#,#" where each '#' marks where a null child would be.

Your task is to determine if the given string is a valid preorder serialization of some binary tree, without actually reconstructing the tree.

The solution uses a stack-based approach. As we process each element from the serialized string:

1. We push each element onto a stack
2. Whenever we see a pattern where the top two elements are '#' (null nodes) and the third element from the top is not '#' (a non-null node), this represents a complete subtree
3. We replace this complete subtree (node with two null children) with a single '#' to represent that entire subtree as processed
4. We repeat this process until no more reductions can be made

After processing all elements, a valid serialization should reduce to exactly one '#' in the stack, representing the entire tree has been properly formed and collapsed.

The key insight is that in a valid preorder traversal, every non-null node must eventually have exactly two children (which could be null), and the pattern of a node followed by two nulls indicates a leaf node that can be collapsed.

Intuition

The key observation is that in a preorder traversal, we visit nodes in the order: root → left subtree → right subtree. When we see a non-null node followed by two '#' symbols, this represents a leaf node (a node with no children). This complete subtree can be thought of as "consumed" or "collapsed" into a single unit.

Think of it like building blocks. When we have a complete subtree (a parent with two null children), we can replace this entire structure with a single '#' to indicate "there was a valid subtree here, and it's now complete."

The pattern recognition works like this:

• [node, '#', '#'] represents a leaf node with value node and two null children
• Once we identify this pattern, the entire subtree rooted at node is complete
• We can replace it with a single '#' to mark this position as a completed subtree

By repeatedly applying this reduction, we're essentially validating the tree from bottom to top. Each reduction confirms that a portion of the tree is structurally valid. If the serialization is correct, all these reductions should eventually collapse the entire tree into a single '#'.

Why does this work? In a valid binary tree:

• Every non-null node must have exactly 2 children positions (filled with either nodes or nulls)
• The number of nulls should be exactly one more than the number of non-null nodes
• The preorder traversal should "consume" all nodes and nulls in a way that forms a complete tree

The stack naturally handles the recursive nature of tree validation - we process nodes in the order they appear, and whenever we can form a complete subtree, we immediately reduce it. If at any point we can't make progress, or if we end with something other than a single '#', the serialization must be invalid.

Solution Approach

The implementation uses a stack to process the serialized string and perform the reduction operations:

1. Split the input string: First, we split the preorder string by commas to get an array of individual nodes and null markers.

2. Process each element with a stack: We iterate through each element and push it onto the stack. The stack helps us keep track of the current state of our validation process.

3. Pattern matching and reduction: After adding each element, we check if the top of the stack matches our reduction pattern:

   • Check if the stack has at least 3 elements
   • Check if the last two elements are both '#' (representing null children)
   • Check if the third-from-last element is NOT '#' (representing a non-null parent node)

   When this pattern [non-null, '#', '#'] is found, it represents a complete leaf node that can be collapsed.

4. Perform the reduction: When we find the pattern, we:

   • Remove the last 3 elements from the stack (stk = stk[:-3])
   • Push a single '#' back onto the stack to represent the entire collapsed subtree

   This reduction continues in a while loop because collapsing one subtree might create a new pattern that can also be collapsed.

5. Final validation: After processing all elements, a valid serialization should result in exactly one element in the stack, and that element should be '#'. This single '#' represents the entire tree collapsed into its simplest form.

The algorithm's correctness relies on the property that in a valid preorder serialization:

• Every complete subtree can be reduced to a single '#'
• The reduction process will eventually collapse the entire tree if and only if the serialization is valid
• Any invalid structure will either leave multiple elements in the stack or won't reduce to a single '#'

Time complexity: O(n) where n is the number of nodes, as each element is pushed and popped at most once. Space complexity: O(n) for the stack in the worst case.

Example Walkthrough

Let's walk through a simple example with the serialization "9,3,#,#,2,#,#".

This represents a tree where:

• Root node is 9
• Left child is 3 (which has two null children)
• Right child is 2 (which has two null children)

Step-by-step process:

1. Initial state: Split the string into ['9', '3', '#', '#', '2', '#', '#'] Stack: []

2. Process '9': Push to stack Stack: ['9']

   • Check pattern: Need 3 elements, only have 1 → no reduction

3. Process '3': Push to stack Stack: ['9', '3']

   • Check pattern: Need 3 elements, only have 2 → no reduction

4. Process first '#': Push to stack Stack: ['9', '3', '#']

   • Check pattern: Have 3 elements, but last two aren't both '#' → no reduction

5. Process second '#': Push to stack Stack: ['9', '3', '#', '#']

   • Check pattern: Last two are '#', third from last '3' is not '#' → MATCH!
   • Reduction: Remove ['3', '#', '#'] and replace with single '#' Stack: ['9', '#']

6. Process '2': Push to stack Stack: ['9', '#', '2']

   • Check pattern: Last element '2' is not '#' → no reduction

7. Process third '#': Push to stack Stack: ['9', '#', '2', '#']

   • Check pattern: Last two aren't both '#' → no reduction

8. Process fourth '#': Push to stack Stack: ['9', '#', '2', '#', '#']

   • Check pattern: Last two are '#', third from last '2' is not '#' → MATCH!
   • Reduction: Remove ['2', '#', '#'] and replace with single '#' Stack: ['9', '#', '#']
   • Check pattern again: Last two are '#', third from last '9' is not '#' → MATCH!
   • Reduction: Remove ['9', '#', '#'] and replace with single '#' Stack: ['#']

9. Final validation: Stack has exactly one element and it's '#' → Valid serialization!

The key insight is that each reduction represents collapsing a complete subtree. Node 3 with its two null children gets reduced first, then node 2 with its null children, and finally the root node 9 with its two now-collapsed subtrees (represented as '#') gets reduced to a single '#'.

Solution Implementation

Time and Space Complexity

The time complexity is O(n) where n is the length of the string preorder. This is because:

• The split(",") operation takes O(n) time to traverse the entire string
• The main loop iterates through each element in the split result, which is at most O(n) elements
• Inside the loop, each element is pushed to the stack once
• The while loop can pop elements, but each element can only be popped once throughout the entire execution
• Therefore, the total number of push and pop operations is bounded by O(n)

The space complexity is O(n) where n is the length of the string preorder. This is because:

• The split(",") operation creates a list that can contain at most O(n) elements (in the worst case, single character nodes separated by commas)
• The stack stk can grow to at most O(n) elements in the worst case before any reduction occurs
• Even though the stack size changes during execution due to the collapsing operation (replacing two "#" and a non-"#" node with a single "#"), the maximum space used is still bounded by O(n)

Common Pitfalls

1. Incorrect Pattern Matching Order

A common mistake is checking the pattern elements in the wrong order or using incorrect indices. Some implementations might accidentally check stack[-3] == "#" instead of stack[-3] != "#", which would look for the wrong pattern and fail to properly validate the tree structure.

Wrong approach:

# Incorrect - looking for wrong pattern
while (len(stack) >= 3 and 
       stack[-1] == "#" and 
       stack[-2] == "#" and 
       stack[-3] == "#"):  # Wrong! Should be != "#"

Solution: Always verify that you're checking for a non-null parent node with two null children: the pattern should be [non-null, '#', '#'] at the top of the stack.

2. Not Handling Edge Cases Properly

The algorithm might fail on edge cases like:

• Empty string ""
• Single null node "#"
• String with only commas ",,,"

Solution: These cases are naturally handled by the split and stack approach, but be aware that:

• "#" is valid (represents an empty tree)
• "" splits to [""] which won't reduce properly and returns False
• Multiple consecutive commas create empty strings in the split array

3. Forgetting to Continue Collapsing After Each Reduction

A critical mistake is performing only one reduction after adding each element, rather than continuing to collapse while the pattern exists. This could leave valid trees unrecognized.

Wrong approach:

for node in preorder.split(","):
    stack.append(node)
    # Only checking once - wrong!
    if (len(stack) >= 3 and 
        stack[-1] == "#" and 
        stack[-2] == "#" and 
        stack[-3] != "#"):
        stack = stack[:-3]
        stack.append("#")

Solution: Use a while loop to keep collapsing as long as the pattern exists, since one reduction might expose another collapsible pattern underneath.

4. Inefficient Stack Manipulation

Using stack = stack[:-3] followed by stack.append("#") creates a new list each time, which could be inefficient for large inputs.

More efficient approach:

while (len(stack) >= 3 and 
       stack[-1] == "#" and 
       stack[-2] == "#" and 
       stack[-3] != "#"):
    stack.pop()  # Remove last element
    stack.pop()  # Remove second-to-last
    stack[-1] = "#"  # Replace third-to-last with "#"

This modifies the stack in-place rather than creating new list slices, improving performance for large trees.