Average of Levels in Binary Tree

Given the root of a binary tree, the task is to return the average value of the nodes on each level in the form of an array. We can solve this problem by performing a level-order traversal and calculating the average of each level’s values.

Algorithm

  1. Initialize: Create an empty result list to store the average of each level.

  2. Level-Order Traversal: Use a queue to perform level-order traversal.

    a. Start: Enqueue the root node to the queue.

    b. Loop: Repeat until the queue is empty: i. Determine the current level size. ii. Initialize a variable to hold the sum of the current level’s values. iii. Dequeue nodes of the current level and add their values to the sum. Also, enqueue their left and right children if they exist. iv. Calculate the average of the current level and append it to the result list.

  3. Return: Return the result list containing the averages.

Code

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from collections import deque

class Solution:
    def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:
        if not root:
            return []

        result = []
        queue = deque([root])

        while queue:
            level_size = len(queue)
            level_sum = 0

            for _ in range(level_size):
                node = queue.popleft()
                level_sum += node.val

                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

            result.append(level_sum / level_size)

        return result

Explanation

  • We perform a level-order traversal using a queue, processing all nodes at each level.
  • For each level, we calculate the sum and divide it by the level’s size to find the average.
  • We append the average of each level to the result list and return it.

Insights

  • Time Complexity: ( O(n) ), where ( n ) is the number of nodes in the tree. We visit each node exactly once.
  • Space Complexity: ( O(w) ), where ( w ) is the maximum width of the tree (i.e., the maximum number of nodes at any level). The space is required for the queue.

This solution efficiently calculates the average of nodes at each level of the binary tree and returns it in the form of an array.

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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 {Float[]}
def average_of_levels(root)
   queue = [root]
   output = []

    while !queue.empty?
        size = queue.size
        sum = 0.0
        size.times do
           node = queue.shift 
            sum += node.val 
            queue << node.left if node.left
            queue << node.right if node.right
        end
        output << sum/size
    end

   return output
end

Identifying Problem Isomorphism

“Average of Levels in Binary Tree” requires you to traverse a binary tree level by level and find the average value of the nodes at each level.

An isomorphic problem to this is “102. Binary Tree Level Order Traversal”. Both problems require level order traversal (also known as breadth-first traversal) of a binary tree. The difference lies in the information you’re collecting: while “Average of Levels in Binary Tree” asks for the average value of nodes at each level, “Binary Tree Level Order Traversal” asks for the list of node values at each level.

“Binary Tree Level Order Traversal” is simpler since it just requires you to collect the node values, while “Average of Levels in Binary Tree” also requires you to calculate the average at each level. However, the underlying traversal logic is the same in both problems.

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def averageOfLevels(self, root: TreeNode) -> List[float]:
	lvlcnt = defaultdict(int)
	lvlsum = defaultdict(int)

	def dfs(node=root, level=0):
		if not node: return
		lvlcnt[level] += 1
		lvlsum[level] += node.val
		dfs(node.left, level+1)
		dfs(node.right, level+1)

	dfs()
	return [lvlsum[i] / lvlcnt[i] for i in range(len(lvlcnt))]

Problem Classification

The problem is in the domain of Trees, and involves concepts of Binary Trees and Breadth-first Search (BFS) traversal of such trees. There’s also a need for Arithmetic operations to calculate averages.

Components of the ‘What’

  1. Binary Tree: The problem involves a binary tree data structure, where each node has at most two children, denoted as left child and right child.

  2. Level-wise Average: The problem requires the calculation of the average value of the nodes for each level in the binary tree.

  3. Array Output: The result of the calculation, i.e., the average values for each level, should be returned in the form of an array.

This problem can be classified as a Tree Traversal problem that requires BFS or level order traversal, since we’re dealing with calculating values (averages) at each level of the tree. The problem also requires Arithmetic Operations to calculate the average of values at each level. It has elements of Computation (calculating averages) and Data Transformation (transforming tree data into an array of averages).

Language Agnostic Coding Drills

  1. Dissect the Code and Identify Each Distinct Concept:

Here are the distinct concepts identified in the given Python code:

a. Importing necessary modules: The code uses defaultdict from the collections module.

b. Recursive function: A depth-first search (DFS) function is implemented using recursion.

c. Dictionary operations: The code updates and fetches values from dictionaries.

d. Conditional (if) statements: If conditions are used to control the flow of the DFS function.

e. Arithmetic operations: The code performs addition and division to calculate level sums and averages.

f. List comprehension: The final average list is constructed using a list comprehension.

  1. Coding Concepts/Drills Order and Difficulty Level:

    a. Importing necessary modules - Easy: This is a simple task of just importing the necessary modules. Knowing what to import comes with knowledge and experience of Python’s standard library.

    b. Conditional (if) statements - Easy: These are fundamental constructs in any programming language.

    c. Arithmetic operations - Easy: Basic mathematical operations such as addition and division.

    d. Dictionary operations - Easy to Intermediate: Requires understanding how dictionaries work, including adding and accessing elements.

    e. Recursive function - Intermediate: This concept is slightly more complex, as recursion requires thinking in terms of function calls and base cases.

    f. List comprehension - Intermediate: While it is a more efficient way to create lists in Python, it might be a little hard for beginners to grasp initially.

  2. Problem-Solving Approach and Step-by-Step Process:

    a. Initialize two dictionaries, one for keeping track of the number of nodes at each level and another for the sum of the node values at each level.

    b. Define a recursive function (DFS) that traverses the tree. For each node, update the corresponding level count and level sum in the dictionaries.

    c. Initiate the DFS from the root of the tree (level 0).

    d. Once the traversal is complete, iterate over the dictionaries to calculate the average value at each level.

    e. Return the list of averages.

Each of the above steps corresponds to the identified coding drills, and by sequentially working through these steps, we are able to solve the problem effectively.

Targeted Drills in Python

  1. Coding Drills for Identified Concepts:

    a. Importing necessary modules:

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    from collections import defaultdict

    b. Conditional (if) statements:

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    x = 5
if x > 0:
    print("Positive number")
else:
    print("Non-positive number")

    c. Arithmetic operations:

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    a = 10
b = 20
sum_ = a + b
avg = sum_ / 2
print(f'Sum: {sum_}, Average: {avg}')

    d. Dictionary operations:

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    d = {}
d['one'] = 1
d['two'] = 2
print(d['one'])
print(d['two'])

    e. Recursive function:

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    def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n-1)
print(factorial(5))

    f. List comprehension:

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    squares = [x**2 for x in range(10)]
print(squares)

  2. Problem-Specific Concepts:

    a. Depth-first Search (DFS): In this problem, a depth-first search is used to traverse the binary tree. DFS visits each node of the tree once, and does so in a depth-first manner.

    Drill:

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    def dfs(node):
    if node is not None:
        print(node.val)
        dfs(node.left)
        dfs(node.right)

  3. Integration of Drills to Solve the Problem:

    a. Start by importing the required modules, in this case defaultdict from collections.

    b. Initialize two defaultdicts, lvlcnt and lvlsum, to keep track of the count and sum of values at each level of the tree.

    c. Define the recursive function dfs to perform a depth-first search on the tree. Inside this function, use conditional statements to check if the node is not None. If not, update lvlcnt and lvlsum with the node’s level and value. Recurse on the left and right children of the node, increasing the level by 1.

    d. Call the dfs function, starting at the root of the tree (level 0).

    e. Finally, use a list comprehension to calculate and return the average value at each level of the tree. Use the lvlsum and lvlcnt defaultdicts to compute these averages.