Taro Logo

Maximum Average Subtree

Medium
Asked by:
Profile picture
Profile picture
6 views
Topics:
TreesRecursion

Given the root of a binary tree, find the maximum average value of any subtree of root.

A subtree of a tree root is a tree consisting of root and all of its descendants.

The average value of a subtree is the sum of its values, divided by the number of nodes.

Return the maximum average value of any subtree of root. Answers within 10-5 of the actual value will be accepted.

Example 1:

Input: root = [5,6,1,null,null,0,8]
Output: 6.00000
Explanation: 
The node with value 6 is the root of a subtree with the value 6.
The node with value 5 is the root of a subtree with the value 5 + 6 + 1 = 12, and the number of nodes is 3, so the average is 12 / 3 = 4.
The node with value 0 is the root of a subtree with the value 0.
The node with value 8 is the root of a subtree with the value 8.
So the maximum average value of any subtree of root is 6 which is the root with value 6.

Example 2:

Input: root = [0,null,1]
Output: 1.00000

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • 0 <= Node.val <= 105

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What is the range of values for the node values in the tree? Can they be negative, zero, or floating-point numbers?
  2. Can the tree be empty or consist of a single node? If so, what should the return value be in these cases?
  3. If multiple subtrees have the same maximum average, can I return any one of them, or is there a tie-breaking condition I should consider?
  4. By 'subtree', do you mean any connected component of the tree, or must the subtree always include the root of the original tree?
  5. Is the input guaranteed to be a valid binary tree, or do I need to handle potential invalid tree structures?

Brute Force Solution

Approach

The brute force approach to finding the maximum average subtree involves examining every single subtree within the entire tree. For each of these subtrees, we'll calculate its average value and compare it against the current maximum average we've found so far. By checking every possible subtree, we guarantee finding the one with the absolute highest average.

Here's how the algorithm would work step-by-step:

  1. Start by considering each node in the tree as the 'root' of a potential subtree.
  2. For a chosen node, calculate the sum of the values of all nodes within the subtree rooted at that node.
  3. Also, count the total number of nodes included in that subtree.
  4. Calculate the average value of the subtree by dividing the sum of the node values by the total number of nodes.
  5. Compare this average to the highest average you've seen so far. If it's higher, remember this new average and the corresponding subtree.
  6. Repeat this process for every single node in the tree.
  7. Once you've gone through all the nodes and their subtrees, the maximum average you've remembered is the final answer.

Code Implementation

class TreeNode:
	def __init__(self, value):
		self.value = value
		self.children = []

def maximum_average_subtree_brute_force(root):
	if not root:
		return None

	maximum_average = float('-inf')
	maximum_average_subtree_root = None

	def calculate_subtree_sum_and_count(node):
		if not node:
			return 0, 0

		subtree_sum = node.value
		subtree_node_count = 1

		# Recursively calculate the sum and count for each child
		for child in node.children:
			child_sum, child_count = calculate_subtree_sum_and_count(child)
			subtree_sum += child_sum
			subtree_node_count += child_count

		return subtree_sum, subtree_node_count

	# Iterate through each node as a potential subtree root
	nodes_to_visit = [root]
	while nodes_to_visit:
		current_node = nodes_to_visit.pop(0)

		# Calculate sum and count for the subtree rooted at current node
		subtree_sum, subtree_node_count = calculate_subtree_sum_and_count(current_node)

		if subtree_node_count > 0:
			subtree_average = subtree_sum / subtree_node_count

			# Compare subtree average with the current maximum average
			if subtree_average > maximum_average:
				maximum_average = subtree_average
				maximum_average_subtree_root = current_node

		# Add children of current node to visit
		for child in current_node.children:
			nodes_to_visit.append(child)

	return maximum_average_subtree_root

Big(O) Analysis

Time Complexity
O(n²)The algorithm iterates through each node of the tree, considering it as the root of a subtree. For each node, it traverses the entire subtree to calculate the sum of node values and the number of nodes. In the worst case, for each of the n nodes, we might visit all the other nodes to compute the sum and count, which takes O(n) time. Therefore, the overall time complexity is approximately n operations performed n times, resulting in roughly n * n operations. This simplifies to O(n²).
Space Complexity
O(N)The brute force approach involves calculating the sum and count of nodes for every subtree. In the worst case, the recursion stack can grow to the height of the tree. For a skewed tree, this height can be equal to the number of nodes, N. Therefore, the auxiliary space used by the recursion stack is proportional to N, resulting in O(N) space complexity.

Optimal Solution

Approach

The key idea is to visit each part of the tree only once to compute information from the bottom up. We want to figure out the average value of each subtree and remember the biggest average we've seen. This avoids recomputing the same averages over and over.

Here's how the algorithm would work step-by-step:

  1. Imagine you are exploring the tree from the bottom up, starting at the leaves.
  2. For each node (which is like a person in a family tree), calculate the total sum of all the values in its subtree (including itself) and the number of nodes in that subtree.
  3. Once you have the sum and the count for a node, calculate the average value for that subtree.
  4. Keep track of the maximum average you have seen so far as you go.
  5. After you have processed the root (the top of the tree), the maximum average you stored is the answer.

Code Implementation

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

class Solution:
    def __init__(self):
        self.maximum_average = float('-inf')

    def find_maximum_average_subtree(self, root):
        self.calculate_subtree_sum_and_size(root)
        return self.maximum_average

    def calculate_subtree_sum_and_size(self, root):
        if not root:
            return 0, 0

        left_sum, left_size = self.calculate_subtree_sum_and_size(root.left)
        right_sum, right_size = self.calculate_subtree_sum_and_size(root.right)

        subtree_sum = root.value + left_sum + right_sum
        subtree_size = 1 + left_size + right_size

        # Calculate the average of the current subtree
        current_average = subtree_sum / subtree_size

        # Update the maximum average seen so far.
        self.maximum_average = max(self.maximum_average, current_average)

        return subtree_sum, subtree_size

Big(O) Analysis

Time Complexity
O(n)The provided solution utilizes a single depth-first traversal of the tree. For a tree with n nodes, each node is visited exactly once to calculate the subtree sum and count. Therefore, the time complexity is directly proportional to the number of nodes in the tree, resulting in O(n) time complexity.
Space Complexity
O(N)The primary driver of space complexity is the recursion depth. The algorithm implicitly uses a call stack to perform a depth-first traversal of the binary tree. In the worst-case scenario (e.g., a skewed tree), the recursion depth can be equal to the number of nodes, N. Therefore, the auxiliary space required for the call stack is proportional to N, leading to O(N) space complexity.

Edge Cases

CaseHow to Handle
Null or empty treeReturn null or an appropriate default value (e.g., null, 0) to indicate no average subtree exists.
Single node treeReturn the single node itself, as it is trivially the subtree with the maximum average.
Tree with all nodes having the same valueThe algorithm should correctly identify any node as a valid maximum average subtree because all subtrees have the same average.
Tree with negative node valuesThe algorithm should correctly handle negative values when calculating subtree sums and averages.
Tree with very large node values leading to potential integer overflowUse long data type or other appropriate data type to store intermediate sums to avoid integer overflow.
Deeply skewed tree (e.g., linked list structure)The recursive algorithm might lead to stack overflow; consider iterative approach if recursion depth becomes a concern.
Floating-point precision issues when calculating averagesConsider using a tolerance value when comparing floating-point averages to account for precision errors.
Very large tree (scalability)Ensure the algorithm has O(n) time complexity, where n is the number of nodes, to handle large trees efficiently, typically achievable with a post-order traversal.