Diameter of Binary Tree

The diameter of a binary tree is the length of the longest path between any two nodes in the tree. This path may or may not pass through the root. To find the diameter, we can perform a depth-first search (DFS) and calculate the depth of the left and right subtrees for each node. The diameter at any node would be the sum of depths of its left and right subtrees.

Python function to find the diameter of a binary tree:

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class Solution:
    def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        self.diameter = 0

        def depth(node):
            if not node:
                return 0
            left_depth = depth(node.left)
            right_depth = depth(node.right)
            # Update the diameter using the depth of the left and right subtrees
            self.diameter = max(self.diameter, left_depth + right_depth)
            # Return the depth of the current subtree
            return 1 + max(left_depth, right_depth)

        depth(root)
        return self.diameter

Explanation:

  1. We use an instance variable self.diameter to keep track of the maximum diameter found so far.
  2. We define a recursive function depth to calculate the depth of a subtree rooted at a given node. This function also updates the diameter using the depths of the left and right subtrees.
  3. Finally, we call the depth function starting from the root and return the calculated diameter.

This function runs in O(N) time complexity, where N is the number of nodes in the tree, as it visits each node exactly once.

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# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} root
# @return {Integer}

    
    1. Work on the example and see how the edges are added
    2. Consider the case where the path does not go through the root
        Example: Consider nodes under 5 that gives us a path of 4. 
    3. Empty tree - return 0
       One node tree - return 0
       Two nodes in a tree - return 1
       Three nodes in a tree - return 2 (left edge + right edge)
            
       Five nodes : 
       7 nodes : 
       
    4. Unit of work (assume we are taking recursive approach)
         - Store the maximum and later update if we find another maximum 
         - left edge + right edge (if the root is part of the longest path)
         
    5. We need to capture the output of the left subproblem and the right subproblem
    6. We must do the work after the recursive calls
    7. What should we return from the recursive call?
        - Return the max of left edge and right edge
    8. Keep updating the longest diameter we have encountered so far
    
=end

def wrapper(root)
    if root.nil?
        return 0
    end
    
    left_edge = wrapper(root.left)
    right_edge = wrapper(root.right)
        
    if root.left
        left_edge = 1 + left_edge
    end
    
    if root.right
        right_edge = 1 + right_edge
    end
    
    # max of left and right     
    @diameter = [@diameter, left_edge + right_edge].max
    
    return [left_edge, right_edge].max
end

def diameter_of_binary_tree(root)
    @diameter = 0
    wrapper(root)
    @diameter
end

“Diameter of Binary Tree” is about finding the length of the longest path between any two nodes in a binary tree, and this path may pass through the root.

This problem is isomorphic to “687. Longest Univalue Path”. The “Longest Univalue Path” problem also requires you to find the length of the longest path in a binary tree, where a path is defined as a sequence of nodes with the same value.

The isomorphism lies in the fact that both problems need to determine the longest path in a binary tree, although the conditions for the path are different. They both use a depth-first search strategy, traversing to the furthest nodes first and then backing up, aggregating the longest path length from the subtrees.

While the two problems share a similar structure and problem-solving approach, they’re not identical. In “543. Diameter of Binary Tree”, any path between two nodes can be considered, while in “687. Longest Univalue Path”, only the path where all nodes have the same value is considered. Thus, the problems are structurally isomorphic, but they’re not exactly the same.

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class Solution:
		def depth(node: Optional[TreeNode]) -> int:
		    """
			To find the maximum depth of a given tree.
			:param node: The (root) node of the tree.
			:returns: The maximum depth of the tree, at the specified node.
			"""
			if not node:
			    return 0  
			left = depth(node.left)  
			right = depth(node.right)  
			return 1 + (left if left > right else right)
  
    def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
	    def depth(node: Optional[TreeNode]) -> int:
		    return 1 + max(depth(node.left), depth(node.right)) if node else 0

		return depth(root.left) + depth(root.right)

Problem Classification

This problem is within the Binary Trees. The main elements to consider in this problem are:

  1. The problem refers to a binary tree, which is a type of data structure where each node has at most two children, commonly designated as the left child and the right child.

  2. The concept of the length of the diameter of the tree is introduced. The diameter of a binary tree is the length of the longest path between any two nodes in a tree, where the length is determined by the number of edges.

  3. It’s specified that the longest path may or may not pass through the root, which adds an additional consideration when searching for the longest path.

This problem can be further classified as a tree traversal problem, as we need to navigate the tree to find the longest path between any two nodes. In terms of complexity, it can be considered a combinatorial optimization problem because there are multiple possible paths, but we need to find the path that maximizes the number of edges.

Additionally, this problem could be seen as an instance of Depth-First Search (DFS) or Dynamic Programming. DFS because we need to traverse the tree deeply to find the longest path, and Dynamic Programming because the longest path to any node can be calculated by optimally combining the longest paths in its left and right subtrees.

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

How did you identify that this problem is a variation of problem?

Language Agnostic Coding Drills

  1. Dissecting the code into distinct concepts:

    • Data structures (Binary Trees): Understanding of binary tree data structure, including its properties, and the TreeNode data type.
    • Function Definition: The process of defining a function and its associated parameters.
    • Recursion: The code employs recursive function calls to traverse the binary tree and calculate its depth.
    • Conditional Statements: The use of if/else statements to control the flow of the function based on whether a node exists.
    • Comparison Operations: Comparing two variables (the depths of left and right subtrees in this case).
    • Mathematical Operations: The use of addition and the max function to calculate the maximum depth of a tree.
    • Nested Function: The concept of defining a function within another function.
    • Optional Parameters: The use of Optional in the function parameters, meaning the parameter could either hold a value or be None.
    • Type Hints: The use of type hints in the function definition to specify expected types of arguments and return type.

  2. Ordering the concepts based on difficulty:

    • Function Definition: The process of defining a function. It is a fundamental concept and is easy to understand.
    • Conditional Statements: They control the flow of the program, an early concept learned in programming.
    • Comparison Operations: Comparing two variables is a fundamental concept in programming.
    • Mathematical Operations: Addition and the max function are simple and widely used operations.
    • Type Hints: They add clarity to code but require some understanding of Python’s typing system.
    • Optional Parameters: Slightly more advanced as it allows for more flexible function arguments.
    • Data structures (Binary Trees): This concept requires understanding of specific data structures, which can be a bit more complex.
    • Recursion: This concept can be difficult to grasp for beginners because it involves the function calling itself.
    • Nested Function: A more advanced concept, understanding how scope works in Python is necessary to correctly use and understand nested functions.

  3. Problem-solving approach:

The solution first defines a helper function depth to calculate the maximum depth of a given tree. This is done recursively, where for a given node, the function computes the maximum depth of its left and right subtrees and adds 1 to it. If the node is None (indicating it’s a leaf node), it returns 0.

For the main problem, the function diameterOfBinaryTree uses the helper function depth to compute the diameter of the tree. This is done by adding the depths of the left and right subtrees of the root. This makes sense because the longest path between any two nodes in a tree is likely to pass through the root, and this path is essentially the sum of the depths of the left and right subtrees.

The problem-solving strategy here involves breaking down the larger problem (calculating the diameter of the tree) into a smaller problem (calculating the depth of a tree), solving the smaller problem using recursion, and then using that solution to solve the larger problem.

Targeted Drills in Python

  1. Python Coding Drills:

    • Function Definition:
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      def greet():
    print("Hello, World!")
greet()
    • Conditional Statements:
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      age = 20
if age >= 18:
    print("You are an adult.")
else:
    print("You are a minor.")
    • Comparison Operations:
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      a = 5
b = 7
if a < b:
    print("a is less than b")
    • Mathematical Operations:
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      a = 5
b = 7
sum = a + b
print("Sum is:", sum)
    • Type Hints:
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      def greet(name: str) -> str:
    return f"Hello, {name}!"
print(greet("World"))
    • Optional Parameters:
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      def greet(name: str = "World"):
    print(f"Hello, {name}!")
greet()
greet("Alice")
    • Data Structures (Binary Trees):
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      class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
    • Recursion:
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      def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n-1)
print(factorial(5))
    • Nested Function:
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      def outer_function(x):
    def inner_function(y):
        return y * y
    return inner_function(x)
print(outer_function(5))

  2. Problem-Specific Concepts:

    The main problem-specific concept for this problem is understanding how to traverse a binary tree to calculate the depth of a tree, which is then used to calculate the diameter of the tree.

    Here’s a simple example of how this concept might be implemented in Python:

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    class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

def maxDepth(node):
    if node is None:
        return 0
    else:
        l_depth = maxDepth(node.left)
        r_depth = maxDepth(node.right)
        return max(l_depth, r_depth) + 1

  3. Integration of Concepts to Solve the Problem:

    Each of the coding drills above contributes to the final solution in different ways. The concept of a binary tree is obviously necessary as the problem involves performing operations on a binary tree. Recursive function calls are used to traverse the binary tree and calculate its depth. Mathematical operations are used to add up the depths and determine the maximum depth, which is used to calculate the diameter of the tree. Nested functions are used to define a helper function within the main function, which makes the code cleaner and more readable.

    To solve the problem, we first define a binary tree Node class and a function to calculate the maximum depth of a tree. Then we define our main function to calculate the diameter of the tree, which internally uses the maxDepth function. We start by calculating the maximum depth of the left and right subtrees, and then return the sum of these two depths, which gives us the diameter of the tree.

title: Diameter of Binary Tree excerpt: This covers how to solve the diameter of a binary tree using recursion tags: return-phase-of-recursion global-variable

Thought Process

  1. Work on the examples and see how the edges are added.

  2. Consider the case where the path does not go through the root. Example: Consider nodes under 5 that give us a path of 4.

  3. Start with simple cases and work your way upto more complex cases.

    Empty tree - return 0 One node tree - return 0 Two nodes in a tree - return 1 Three nodes in a tree - return 2 (left edge + right edge)

    Five nodes : What should we return? 7 nodes : What should we return?

  4. Unit of work (assume we are taking recursive approach)

    • Store the maximum and later update if we find another maximum
    • left edge + right edge (if the root is part of the longest path)

  5. We need to capture the output of the left subproblem and the right subproblem.

  6. We must do the work after the recursive calls.

  7. What should we return from the recursive call?

    • Return the max of left edge and right edge.

  8. Keep updating the longest diameter we have encountered so far.

Working Code

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def wrapper(root)
  return 0 if root.nil?
    
  left = wrapper(root.left)
  right = wrapper(root.right)
    
  left_edge = 0
  right_edge = 0
    
  if root.left
    left_edge = 1 + left
  end
    
  if root.right
    right_edge = 1 + right
  end
    
  # max of left and right     
  @diameter = [@diameter, left_edge + right_edge].max
    
  return [left_edge, right_edge].max
end

def diameter_of_binary_tree(root)
  @diameter = 0
  wrapper(root)
  @diameter
end

Refactored Solution

We can remove the unnecessary variables from the recursive method.

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def wrapper(root)
  if root.nil?
    return 0
  end
    
  left_edge = wrapper(root.left)
  right_edge = wrapper(root.right)
        
  if root.left
    left_edge = 1 + left_edge
  end
    
  if root.right
    right_edge = 1 + right_edge
  end
    
  # max of left and right     
  @diameter = [@diameter, left_edge + right_edge].max
    
  return [left_edge, right_edge].max
end

def diameter_of_binary_tree(root)
  @diameter = 0
  wrapper(root)
  @diameter
end