Problem Description

Given the root of a binary tree, you need to find the maximum average value among all possible subtrees in that tree.

A subtree is defined as any node in the tree along with all of its descendants. Every node by itself (with no children) also counts as a subtree.

The average value of a subtree is calculated by taking the sum of all node values in that subtree and dividing it by the total number of nodes in that subtree.

For example, if a subtree contains nodes with values [5, 6, 1], the average would be (5 + 6 + 1) / 3 = 4.

Your task is to examine all possible subtrees in the given binary tree and return the highest average value found. The answer will be accepted if it's within 10^-5 of the actual answer (to account for floating-point precision).

The solution uses a depth-first search (DFS) approach where for each node, it calculates:

1. The sum of all values in the subtree rooted at that node
2. The count of nodes in that subtree
3. The average value (sum/count) for comparison with the current maximum

The function dfs(root) returns a tuple (sum, count) for each subtree, allowing the parent nodes to aggregate these values. As we traverse each node, we calculate its subtree's average and update the global maximum if a higher average is found.

Flowchart Walkthrough

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

Is it a graph?

• Yes: The problem involves a binary tree, which is a special type of graph (specifically, a connected acyclic graph).

Is it a tree?

• Yes: The problem explicitly states we're working with a binary tree structure where we need to examine all subtrees.

DFS

• We arrive at DFS: Since we're dealing with a tree and need to explore all subtrees (every node and its descendants), DFS is the natural choice.

Why DFS is appropriate here:

• We need to visit every node in the tree to consider it as a potential subtree root
• For each node, we need information from its children before we can calculate the subtree's average (post-order traversal)
• DFS allows us to recursively calculate the sum and count of nodes in each subtree from bottom to top
• The recursive nature of DFS perfectly matches the recursive structure of the tree

Conclusion: The flowchart correctly leads us to DFS for this problem. The DFS pattern is ideal because we need to:

1. Traverse the entire tree to examine all possible subtrees
2. Aggregate information (sum and count) from child nodes before processing parent nodes
3. Calculate averages for each subtree during the traversal
4. Track the maximum average found across all subtrees

Intuition

To find the maximum average among all subtrees, we need to calculate the average for every possible subtree in the tree. A key observation is that to calculate the average of any subtree, we need two pieces of information: the sum of all node values in that subtree and the count of nodes in that subtree.

Think about how we can get this information efficiently. If we're at a node, we can calculate its subtree's sum and count if we know:

• The sum and count from its left subtree
• The sum and count from its right subtree
• Its own value (which adds 1 to the count and node.val to the sum)

This naturally suggests a bottom-up approach where we first process the children before processing the parent. This is exactly what post-order DFS gives us.

For each node during our traversal:

1. We recursively get (sum, count) from the left child
2. We recursively get (sum, count) from the right child
3. We calculate the current subtree's total sum: current_sum = node.val + left_sum + right_sum
4. We calculate the current subtree's total count: current_count = 1 + left_count + right_count
5. We compute the average: average = current_sum / current_count
6. We update our global maximum if this average is larger

The beauty of this approach is that we calculate each subtree's average exactly once while traversing the tree, making it efficient with O(n) time complexity. By returning (sum, count) from each recursive call, parent nodes can easily aggregate the information from their children without recalculating anything.

The base case is straightforward: a null node contributes 0 to both sum and count, represented as (0, 0).

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

Solution Approach

The implementation uses a recursive DFS approach with a helper function that returns two values for each subtree: the sum of node values and the count of nodes.

Implementation Details:

1. Main Function Setup:

   • Initialize a variable ans to track the maximum average found so far
   • Call the DFS helper function starting from the root
   • Return the final maximum average

2. DFS Helper Function dfs(root):

   • Base Case: If the current node is None, return (0, 0) representing zero sum and zero count

   • Recursive Calls:

     • Call dfs(root.left) to get the sum and count from the left subtree: ls, ln
     • Call dfs(root.right) to get the sum and count from the right subtree: rs, rn

   • Calculate Current Subtree Values:

     • Sum of current subtree: s = root.val + ls + rs
     • Count of nodes in current subtree: n = 1 + ln + rn
     • Average of current subtree: s / n

   • Update Maximum:

     • Compare the current subtree's average with the global maximum
     • Update using: ans = max(ans, s / n)

   • Return Values:

     • Return (s, n) so parent nodes can use these values in their calculations

Key Design Choices:

• Using nonlocal: The ans variable is declared as nonlocal inside the DFS function to allow modification of the outer scope variable, enabling us to track the global maximum across all recursive calls

• Post-order Traversal: The algorithm processes children before parents (left → right → root), ensuring we have all necessary information from descendants before calculating a node's subtree average

• Single Pass: Each node is visited exactly once, and its subtree information is calculated only once, achieving O(n) time complexity

Time and Space Complexity:

• Time: O(n) where n is the number of nodes, as we visit each node exactly once
• Space: O(h) where h is the height of the tree, due to the recursive call stack

Example Walkthrough

Let's walk through a concrete example to understand how the solution works.

Consider this binary tree:

       5
      / \
     6   1
    /
   2

We want to find the maximum average among all possible subtrees.

Step-by-step DFS traversal (post-order):

1. Start at root (5), but first we need to process its children.

2. Go to left child (6), but first process its children.

3. Go to node 2 (leaf):

   • Left child: None → returns (0, 0)
   • Right child: None → returns (0, 0)
   • Current sum: 2 + 0 + 0 = 2
   • Current count: 1 + 0 + 0 = 1
   • Average: 2/1 = 2.0
   • Update max: ans = 2.0
   • Return: (2, 1)

4. Back to node 6:

   • Left child returned: (2, 1)
   • Right child: None → returns (0, 0)
   • Current sum: 6 + 2 + 0 = 8
   • Current count: 1 + 1 + 0 = 2
   • Average: 8/2 = 4.0
   • Update max: ans = 4.0 (since 4.0 > 2.0)
   • Return: (8, 2)

5. Go to right child (1):

   • Left child: None → returns (0, 0)
   • Right child: None → returns (0, 0)
   • Current sum: 1 + 0 + 0 = 1
   • Current count: 1 + 0 + 0 = 1
   • Average: 1/1 = 1.0
   • Max stays: ans = 4.0 (since 1.0 < 4.0)
   • Return: (1, 1)

6. Back to root (5):

   • Left child returned: (8, 2)
   • Right child returned: (1, 1)
   • Current sum: 5 + 8 + 1 = 14
   • Current count: 1 + 2 + 1 = 4
   • Average: 14/4 = 3.5
   • Max stays: ans = 4.0 (since 3.5 < 4.0)
   • Return: (14, 4)

Result: The maximum average is 4.0, found in the subtree rooted at node 6 (containing nodes 6 and 2).

All subtrees and their averages:

• Subtree rooted at 2: avg = 2/1 = 2.0
• Subtree rooted at 6: avg = 8/2 = 4.0 ✓ (maximum)
• Subtree rooted at 1: avg = 1/1 = 1.0
• Subtree rooted at 5: avg = 14/4 = 3.5

The algorithm efficiently computes each subtree's average exactly once by propagating sum and count values up the tree, allowing parent nodes to reuse calculations from their children.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm performs a depth-first search (DFS) traversal of the binary tree. Each node is visited exactly once during the traversal. At each node, the operations performed are:

• Recursive calls to left and right children
• Constant time arithmetic operations (addition for sum, increment for count)
• Constant time comparison for updating the maximum average

Since we visit each of the n nodes exactly once and perform O(1) operations at each node, the overall time complexity is O(n).

Space Complexity: O(n)

The space complexity comes from the recursive call stack used during the DFS traversal. In the worst case, the tree could be completely unbalanced (essentially a linked list), where the depth of recursion would be n. This would require O(n) space on the call stack.

In the best case of a perfectly balanced tree, the recursion depth would be O(log n), but since we consider worst-case complexity, the space complexity is O(n).

Additionally, the algorithm uses only a constant amount of extra space for variables (ans, s, n, ls, ln, rs, rn), which doesn't affect the overall space complexity.

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

Common Pitfalls

1. Integer Division Instead of Float Division

A critical mistake is performing integer division when calculating the average, which truncates the decimal part and leads to incorrect results.

Incorrect Implementation:

# In Python 2 or with integer division
current_average = subtree_sum // subtree_count  # Wrong! Loses precision

Correct Implementation:

# Use float division to maintain precision
current_average = subtree_sum / subtree_count  # Correct
# Or explicitly convert to float
current_average = float(subtree_sum) / float(subtree_count)

2. Not Handling Single Node Trees Properly

Forgetting that a single node (leaf) is also a valid subtree can cause incorrect initialization or comparison logic.

Incorrect Implementation:

def maximumAverageSubtree(self, root):
    if not root:
        return 0
  
    max_average = float('-inf')  # Could miss single negative node
    # ... rest of code

Correct Implementation:

def maximumAverageSubtree(self, root):
    max_average = 0.0  # Or float('-inf') if negative values are possible
  
    def dfs(node):
        if node is None:
            return 0, 0
        # ... handle all nodes including leaves

3. Forgetting the nonlocal Declaration

Without declaring max_average as nonlocal, the code will either create a local variable or raise an UnboundLocalError when trying to modify it.

Incorrect Implementation:

def maximumAverageSubtree(self, root):
    max_average = 0.0
  
    def dfs(node):
        # ... calculate current_average
        max_average = max(max_average, current_average)  # UnboundLocalError!
        return subtree_sum, subtree_count

Correct Implementation:

def maximumAverageSubtree(self, root):
    max_average = 0.0
  
    def dfs(node):
        nonlocal max_average  # Essential declaration
        # ... calculate current_average
        max_average = max(max_average, current_average)
        return subtree_sum, subtree_count

4. Incorrect Order of Operations

Calculating the average before having complete subtree information leads to wrong results.

Incorrect Implementation:

def dfs(node):
    if node is None:
        return 0, 0
  
    # Wrong: Calculating average before getting subtree info
    current_average = node.val / 1
    max_average = max(max_average, current_average)
  
    left_sum, left_count = dfs(node.left)
    right_sum, right_count = dfs(node.right)
    # Missing the actual subtree average calculation

Correct Implementation:

def dfs(node):
    if node is None:
        return 0, 0
  
    # First, get complete subtree information
    left_sum, left_count = dfs(node.left)
    right_sum, right_count = dfs(node.right)
  
    # Then calculate the complete subtree values
    subtree_sum = node.val + left_sum + right_sum
    subtree_count = 1 + left_count + right_count
  
    # Finally, calculate and compare average
    current_average = subtree_sum / subtree_count
    nonlocal max_average
    max_average = max(max_average, current_average)
  
    return subtree_sum, subtree_count

5. Returning Wrong Values from DFS

The helper function must return both sum and count for parent calculations. Returning only the average breaks the recursive logic.

Incorrect Implementation:

def dfs(node):
    # ... calculate everything
    current_average = subtree_sum / subtree_count
    return current_average  # Wrong! Parent can't use this

Correct Implementation:

def dfs(node):
    # ... calculate everything
    current_average = subtree_sum / subtree_count
    nonlocal max_average
    max_average = max(max_average, current_average)
    return subtree_sum, subtree_count  # Return sum and count for parent