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:
[1, 104]
.0 <= Node.val <= 105
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:
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:
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
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:
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
Case | How to Handle |
---|---|
Null or empty tree | Return null or an appropriate default value (e.g., null, 0) to indicate no average subtree exists. |
Single node tree | Return the single node itself, as it is trivially the subtree with the maximum average. |
Tree with all nodes having the same value | The algorithm should correctly identify any node as a valid maximum average subtree because all subtrees have the same average. |
Tree with negative node values | The algorithm should correctly handle negative values when calculating subtree sums and averages. |
Tree with very large node values leading to potential integer overflow | Use 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 averages | Consider 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. |