Construct String from Binary Tree

We can use a recursive function to construct the string for the given binary tree. Here’s a step-by-step explanation:

  1. Base Case: If the node is None, return an empty string.
  2. Process Current Node: Include the current node’s value in the result.
  3. Process Left Subtree: If there is a left child, recursively process it and include the result in parentheses. If there is no left child but there is a right child, include an empty set of parentheses to ensure the correct one-to-one mapping.
  4. Process Right Subtree: If there is a right child, recursively process it and include the result in parentheses.
  5. Combine Results: Combine the results of the current node and its children.

Here’s the code:

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class Solution:
    def tree2str(self, root: Optional[TreeNode]) -> str:
        # Base case: if the node is None, return an empty string
        if root is None:
            return ""

        # Process the current node's value
        result = str(root.val)

        # Process the left subtree
        left_subtree = self.tree2str(root.left)
        if left_subtree or root.right:
            result += f"({left_subtree})"

        # Process the right subtree
        right_subtree = self.tree2str(root.right)
        if right_subtree:
            result += f"({right_subtree})"

        return result

Explanation:

  • The function processes the tree in preorder traversal, handling the current node first, then the left subtree, and finally the right subtree.
  • The use of parentheses ensures that the string maintains the original tree’s structure, omitting unnecessary empty parentheses when possible.

The time complexity of this approach is O(n), where n is the number of nodes in the tree, and the space complexity is also O(n) considering the function call stack.

Identifying Problem Isomorphism

“Construct String from Binary Tree” can be mapped to “Binary Tree Inorder Traversal”.

“Construct String from Binary Tree” involves creating a string representation of a binary tree where each node is represented by its value, and the left and right children are represented by parentheses containing their string representations.

“Binary Tree Inorder Traversal” also requires you to traverse the binary tree. The only difference is that in “Binary Tree Inorder Traversal”, instead of creating a string representation, you are creating a list of node values. In both cases, you’re performing a traversal of the tree and processing the node values.

While these problems are not exact isomorphisms of each other, the underlying tree traversal operation is the same. The main difference is the way in which the output is represented.

“Binary Tree Inorder Traversal” is a simpler problem since it only involves collecting the node values during traversal, while “Construct String from Binary Tree” additionally requires constructing a formatted string.

“Construct String from Binary Tree” involves concepts like binary tree traversal and string manipulation. Here are 10 problems to prepare for it:

  1. 104. Maximum Depth of Binary Tree: This problem involves a simple traversal of the binary tree, which is a useful basic skill for dealing with binary trees.

  2. 108. Convert Sorted Array to Binary Search Tree: This problem requires creating a binary tree from a sorted array, which is useful for understanding how binary trees are structured.

  3. 110. Balanced Binary Tree: This problem asks you to judge whether a binary tree is balanced or not, requiring you to traverse and understand the structure of the tree.

  4. 226. Invert Binary Tree: Inverting a binary tree is a basic operation that can help you understand the structure of binary trees.

  5. 257. Binary Tree Paths: This problem involves traversing a binary tree and doing some manipulation with the results, which can be a good practice for similar tasks in the target problem.

  6. 283. Move Zeroes: This problem involves array manipulation and can be a good way to practice basic string or array manipulation skills.

  7. 344. Reverse String: This problem asks you to reverse a string, which is a basic string operation that you might need in the target problem.

  8. 404. Sum of Left Leaves: This problem asks you to sum up the left leaves of a binary tree, requiring you to distinguish left and right children of a node.

  9. 543. Diameter of Binary Tree: This problem involves both depth-first traversal and some calculations based on the traversal results.

  10. 617. Merge Two Binary Trees: This problem requires you to traverse two trees simultaneously and could be a good exercise for more complex binary tree operations.

These cover a range of binary tree operations, from creation to traversal to manipulation, which will be useful when dealing with binary tree problems like the target problem.

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

# Time Complexity: O(N) where N is the number of the nodes in the tree
# Space Complexity: O(H) where H is the height of the tree. 
# In worse case, H can be N when it is a left skewed binary tree / right skewed binary tree
class Solution:
    def tree2str(self, root: Optional[TreeNode]) -> str:
        if root is None:
            return ''

        s = str(root.val)

        if root.left is None and root.right is None:
            s += ''

        if root.left:
            s += '({})'.format(self.tree2str(root.left))

        if root.left is None and root.right:
            s += '()'

        if root.right:
            s += '({})'.format(self.tree2str(root.right))
        return s

Problem Classification

This problem falls under the Trees and Binary Trees. Additionally, it involves String Manipulation.

What components:

  1. Binary Tree: The problem provides a binary tree as input. A binary tree is a tree data structure in which each node has at most two children, referred to as the left child and the right child.

  2. Root of a Binary Tree: The problem specifically mentions the root of a binary tree. The root of a tree is the node where no nodes point to it, and every other node can be reached from it.

  3. Preorder Traversal: The problem requires the construction of a string using a preorder traversal of the binary tree. Preorder traversal is a type of depth-first traversal where each node is visited in the order: Root node, left subtree, right subtree.

  4. String Construction: The task is to construct a string consisting of parentheses and integers derived from the binary tree.

  5. Omission of Empty Parenthesis Pairs: The problem specifies a constraint that any parenthesis pair that does not affect the mapping between the string and the binary tree should be omitted.

This problem can be classified as a String Construction Problem using Tree Traversal (Preorder). It involves traversing the tree in a specific order (preorder) and converting the data into a specific format (a string). It tests the understanding of binary trees, tree traversal techniques, and string manipulation.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

Language Agnostic Coding Drills

  1. Dissect the code and identify each distinct concept:

    a. Understanding Data Structures: The code uses a binary tree data structure, which is a fundamental concept.

    b. Null/None Checking: The code checks if a node is None (or null in other languages), a common operation when working with nodes in a tree.

    c. Value Accessing from Object/Struct: The code accesses the ‘val’, ’left’, and ‘right’ properties of the TreeNode object.

    d. String Concatenation/Formatting: The code combines string literals and variables to construct a new string.

    e. Conditional Checks (if statements): The code includes various if statements to perform different actions based on the properties of the TreeNode.

    f. Recursive Function Calls: The function tree2str calls itself recursively, which is a common technique in tree-based algorithms.

  2. Coding concepts in order of increasing difficulty:

    a. Understanding Data Structures - Basic level: Understanding how data structures, particularly binary trees, work is foundational for the problem.

    b. Null/None Checking - Basic level: It’s a basic programming concept necessary to prevent null pointer exceptions or similar errors.

    c. Value Accessing from Object/Struct - Basic level: It’s a common operation in most object-oriented and structured programming languages.

    d. String Concatenation/Formatting - Intermediate level: It requires understanding how to manipulate strings and format them using variables.

    e. Conditional Checks (if statements) - Intermediate level: It requires understanding of logic and control flow, but is a fundamental concept in most programming languages.

    f. Recursive Function Calls - Advanced level: It requires a good understanding of recursion, stack memory, and function call mechanics.

  3. Problem-solving approach:

    The problem involves converting a binary tree into a specific string format using a preorder traversal. The approach can be broken down as follows:

    a. Begin at the root of the tree. Check if the root is None. If it is, return an empty string as there’s nothing to convert.

    b. Convert the value of the root node into a string and assign it to s.

    c. Check if the left and right children of the root node are None. If they are, no further steps are needed and we can move on.

    d. If the left child exists, recursively call the tree2str function for the left child and add its result to s in parentheses.

    e. If the right child exists but the left child doesn’t, add an empty set of parentheses to s.

    f. If the right child exists, recursively call the tree2str function for the right child and add its result to s in parentheses.

    g. Finally, return s. The recursion will continue for all nodes of the tree until the entire tree has been processed and the desired string format has been achieved.

Targeted Drills in Python

  1. Python-based coding drills for each identified concept:

    a. Understanding Data Structures (Binary Trees):

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    class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

    b. Null/None Checking:

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    x = None
if x is None:
    print("x is None!")

    c. Value Accessing from Object/Struct:

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    class MyClass:
    def __init__(self, x):
        self.value = x

my_obj = MyClass(5)
print(my_obj.value)  # Prints: 5

    d. String Concatenation/Formatting:

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    s = "Hello"
s += " World!"  # String concatenation
print(s)  # Prints: Hello World!

name = "John"
greeting = "Hello, {}!".format(name)  # String formatting
print(greeting)  # Prints: Hello, John!

    e. Conditional Checks (if statements):

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    x = 10
if x > 5:
    print("x is greater than 5")  # This will be printed
if x < 5:
    print("x is less than 5")

    f. Recursive Function Calls:

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    def factorial(n):
    if n == 0:
        return 1
    else:
        return n * factorial(n-1)

print(factorial(5))  # Prints: 120

  2. Problem-specific concepts:

    a. Tree Traversal (Preorder): Traversing a binary tree is a key concept for this problem. In particular, we need to understand preorder traversal, which visits the root first, followed by the left child, and finally the right child. The recursion drill is directly related to this concept.

    b. Parentheses Addition Based on Tree Structure: The problem-specific rule of omitting parentheses for non-existing subtrees needs to be understood and drilled.

  3. Integration of drills to solve the initial problem:

    Starting with the creation of a binary tree, we use the TreeNode class. We would then create the tree with some values. Next, we would need to implement a function to convert the tree into a string. This function should check if the node is None, then obtain the node’s value, and then recursively apply the function to the left and right children, according to the rules given in the problem statement. At each step, the function would use string concatenation and formatting to add parentheses as necessary. Finally, the result would be returned. The understanding of if conditions would be necessary to implement the rules correctly. And recursion would be crucial to traverse the tree and build the string.