Problem Description

You are given a string representation of a binary tree and need to construct the actual binary tree from it.

The string follows these rules:

• Each node is represented by an integer value
• Child nodes (if they exist) are enclosed in parentheses immediately after their parent's value
• The first pair of parentheses after a node contains its left child
• The second pair of parentheses (if present) contains its right child
• The pattern is recursive - each child follows the same structure

For example:

• "4(2(3)(1))(6(5))" represents a tree where:
  • Root node has value 4
  • Left child has value 2 (with its own left child 3 and right child 1)
  • Right child has value 6 (with only a left child 5)

The parsing algorithm works by:

1. Finding the root value (the number before any parentheses)
2. Using parenthesis counting to identify where the left subtree ends
3. Recursively parsing the left subtree content
4. If more content exists after the left subtree, recursively parsing the right subtree
5. Building the tree by connecting the parsed nodes

The solution tracks opening and closing parentheses to determine the boundaries of each subtree. When the count of ( and ) balances back to zero, we've found the end of a complete subtree expression. The first balanced section becomes the left child, and any remaining balanced section becomes the right child.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: While this problem involves a binary tree (which is a special type of graph), we need to construct it from a string representation. The tree structure with nodes and parent-child relationships makes this a graph problem.

Is it a tree?

• Yes: The problem explicitly asks us to construct a binary tree. Each node has at most two children (left and right), and there's a clear hierarchical structure with a root node.

DFS

• Yes: We arrive at DFS as the solution approach.

Conclusion: The flowchart suggests using a Depth-First Search (DFS) approach for this problem.

This makes sense because:

1. We need to recursively parse nested parentheses structures in the string
2. Each time we encounter a node value followed by parentheses, we need to dive deep into those parentheses to construct the subtrees before moving on
3. The recursive nature of the problem (a tree contains smaller trees) naturally fits the DFS pattern
4. We process each node and its entire subtree completely before backtracking to process siblings, which is the essence of depth-first traversal

The DFS approach allows us to:

• Parse the root value
• Recursively construct the left subtree from the first pair of parentheses
• Recursively construct the right subtree from the second pair of parentheses (if it exists)
• Build the tree bottom-up by connecting child nodes to their parents as we return from recursive calls

Intuition

When we look at the string representation like "4(2(3)(1))(6(5))", we can see a recursive pattern - the entire string follows the same structure as any substring within parentheses. This immediately suggests a recursive approach.

The key insight is that parentheses act as delimiters that define the boundaries of subtrees. Just like how we use parentheses in mathematical expressions to group operations, here they group tree nodes. Each balanced pair of parentheses contains a complete subtree that follows the exact same format as the original string.

To parse this string, we need to:

1. Extract the root value (the number before any parentheses)
2. Find where the left subtree ends (when parentheses balance)
3. Recursively solve the same problem for what's inside those parentheses

The challenge is determining where one subtree ends and another begins. Consider "4(2(3)(1))(6(5))" - after reading 4(, how do we know the left subtree is 2(3)(1) and not just 2(? We need to count parentheses: when we've seen equal numbers of opening and closing parentheses, we've found a complete subtree.

This counting mechanism works like a stack - each ( pushes us deeper into nested structures, and each ) pops us back out. When our counter returns to zero, we know we've completely exited the current subtree.

The recursive DFS pattern emerges naturally because:

• We process the current node first (the root value)
• Then dive deep into the left child's substring
• Finally dive into the right child's substring (if it exists)
• Each recursive call handles a self-contained subtree string

This approach mirrors how we mentally parse nested structures - we identify the top-level components first, then recursively break down each component using the same rules.

Learn more about Stack, Tree, Depth-First Search and Binary Tree patterns.

Solution Approach

The implementation uses a recursive DFS function to parse the string and build the tree. Let's walk through the key components:

Base Case Handling: The function first checks if the string is empty - if so, it returns None since there's no node to create. This handles cases where a node has no children.

Root Value Extraction: We find the first occurrence of ( using s.find('('). If no parenthesis is found (p == -1), the entire string is just a number, so we create and return a leaf node with TreeNode(int(s)).

If parentheses exist, we extract the root value as everything before the first (: int(s[:p]).

Parenthesis Counting Algorithm: The core parsing logic uses a counter-based approach:

• Initialize cnt = 0 and start = p (position of first ()
• Iterate through the string from position p
• Increment cnt for each (
• Decrement cnt for each )
• When cnt reaches 0, we've found a complete balanced subtree

Identifying Left and Right Subtrees: The algorithm uses the start variable to track which subtree we're parsing:

• When start == p (first time cnt reaches 0), we've found the left subtree
  • Extract substring: s[start + 1 : i] (content between parentheses)
  • Recursively parse it: root.left = dfs(s[start + 1 : i])
  • Update start = i + 1 to point past this subtree
• When start != p (second time cnt reaches 0), we've found the right subtree
  • Extract and parse similarly: root.right = dfs(s[start + 1 : i])

String Slicing: The solution uses Python's string slicing to extract substrings:

• s[:p] - gets the root value
• s[start + 1 : i] - gets content between parentheses (excluding the parentheses themselves)

Tree Construction: The tree is built bottom-up through the recursive calls:

1. Deepest recursive calls create leaf nodes
2. As recursion unwinds, parent nodes are created and connected to their children
3. Finally, the root node with all descendants is returned

The time complexity is O(n) where n is the length of the string, as we process each character once. The space complexity is O(h) for the recursion stack, where h is the height of the resulting tree.

Example Walkthrough

Let's trace through the algorithm with a simple example: "1(2)(3(4))"

This represents a tree where:

• Root = 1
• Left child = 2 (leaf)
• Right child = 3 with left child = 4

Step 1: Initial call with "1(2)(3(4))"

• Find first ( at position 1
• Extract root value: s[:1] = "1", create root = TreeNode(1)
• Start counting from position 1

Step 2: Find left subtree

• start = 1, cnt = 0
• Position 1: ( → cnt = 1
• Position 2: 2 → skip
• Position 3: ) → cnt = 0 ✓ Found complete subtree!
• Left subtree string: s[2:3] = "2"
• Recursively call dfs("2")
  • No parentheses found, return TreeNode(2)
• Set root.left = TreeNode(2)
• Update start = 4

Step 3: Find right subtree

• Continue from position 4, cnt = 0
• Position 4: ( → cnt = 1
• Position 5: 3 → skip
• Position 6: ( → cnt = 2
• Position 7: 4 → skip
• Position 8: ) → cnt = 1
• Position 9: ) → cnt = 0 ✓ Found complete subtree!
• Right subtree string: s[5:9] = "3(4)"
• Recursively call dfs("3(4)")

Step 4: Parse right subtree "3(4)"

• Find first ( at position 1
• Extract root value: "3", create node = TreeNode(3)
• Count parentheses starting at position 1:
  • Position 1: ( → cnt = 1
  • Position 2: 4 → skip
  • Position 3: ) → cnt = 0 ✓
• Left subtree string: s[2:3] = "4"
• Recursively call dfs("4")
  • No parentheses, return TreeNode(4)
• Set node.left = TreeNode(4)
• No more content, so no right child
• Return TreeNode(3) with left child 4

Step 5: Complete the tree

• Set root.right = TreeNode(3) (which has left child 4)
• Return the complete tree

Final tree structure:

    1
   / \
  2   3
     /
    4

The parenthesis counting ensures we correctly identify (2) as the complete left subtree and (3(4)) as the complete right subtree, even though the right subtree contains nested parentheses.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²) where n is the length of the input string.

The analysis breaks down as follows:

• The function traverses each character in the string to find matching parentheses, which takes O(n) time in the worst case
• String slicing operations s[start + 1 : i] create new substrings, which takes O(k) time where k is the length of the substring
• In the worst case, we process every character of the string through recursive calls, and each recursive call involves creating substrings
• The total work done across all recursive calls involves processing each character and potentially copying it during substring operations, leading to O(n²) complexity

Space Complexity: O(n) where n is the length of the input string.

The space complexity consists of:

• Recursion stack: The maximum depth of recursion equals the height of the tree, which in the worst case (skewed tree) is O(n)
• Substring creation: Each recursive call creates new substrings from the original string. While individual substrings are created, the total space used for all substrings across all recursive calls is bounded by O(n) since we're essentially partitioning the original string
• Tree nodes: Creating O(n) tree nodes, but this is typically not counted as auxiliary space since it's the output
• Overall auxiliary space complexity is O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Parsing of Negative Numbers

The current implementation assumes all node values are non-negative integers. When the input contains negative numbers like "-4(2)(3)", the code will fail because it doesn't handle the minus sign properly when finding the first parenthesis or extracting the root value.

Why it fails: The find('(') method locates the first parenthesis, but when extracting s[:first_paren_index], if there's a negative number, int() conversion works fine. However, the real issue arises in more complex parsing scenarios where negative numbers can be confused with operators or delimiters.

Solution:

def str2tree(self, s: str) -> TreeNode:
    def build_tree(tree_string: str) -> TreeNode:
        if not tree_string:
            return None
      
        # Find where the number ends (could be negative)
        i = 0
        # Handle negative sign
        if i < len(tree_string) and tree_string[i] == '-':
            i += 1
        # Parse all digits
        while i < len(tree_string) and tree_string[i].isdigit():
            i += 1
      
        # If we've consumed the entire string, it's a leaf node
        if i == len(tree_string):
            return TreeNode(int(tree_string))
      
        # Extract root value and create node
        root = TreeNode(int(tree_string[:i]))
      
        # Now i points to the first '('
        start_position = i
        parenthesis_count = 0
      
        # Continue with parenthesis matching logic...
        for current_index in range(i, len(tree_string)):
            if tree_string[current_index] == '(':
                parenthesis_count += 1
            elif tree_string[current_index] == ')':
                parenthesis_count -= 1
          
            if parenthesis_count == 0:
                if start_position == i:
                    root.left = build_tree(tree_string[start_position + 1 : current_index])
                    start_position = current_index + 1
                else:
                    root.right = build_tree(tree_string[start_position + 1 : current_index])
      
        return root
  
    return build_tree(s)

2. Edge Case: Single Child Nodes

Another pitfall is not properly handling cases where a node has only a right child but no left child. According to the problem description, the first pair of parentheses always represents the left child. If a node has only a right child, it should be represented as "val()(right_child)" with empty parentheses for the missing left child.

Example that might cause issues: "1()(2)" - node 1 with no left child and right child 2.

The current code handles this correctly because when it encounters "()", the recursive call with an empty string returns None, which is the desired behavior. However, developers might overlook testing this case or might try to "optimize" by skipping empty parentheses, which would break the logic.

3. Stack Overflow with Deep Trees

For extremely deep trees, the recursive approach could cause a stack overflow due to Python's recursion limit (default is around 1000).

Solution for deep trees: Use an iterative approach with an explicit stack, or increase the recursion limit:

import sys
sys.setrecursionlimit(10000)  # Increase if needed

However, be cautious with this approach as it could lead to actual stack overflow at the system level for very large inputs.