1120. Maximum Average Subtree

Problem Description

The problem provides us with the root node of a binary tree and asks us to calculate the maximum average value of any subtree within that binary tree. Here, a subtree can be any node and all of its descendants—it could be the entire tree, a leaf node, or any other set of connected nodes within the tree.

The average value of a subtree is the sum of all node values in that subtree divided by the number of nodes it contains. Since we are looking for the maximum average, we are concerned with finding the subtree that has the highest average value compared to all other subtrees in the binary tree.

Intuition

The intuitive approach to solving this problem is to explore each subtree, calculate its sum and node count, and then use these to compute the average value for that particular subtree. To achieve this, we'll need to perform a traversal of the binary tree.

A depth-first search (DFS) traversal is suitable for this task because it allows us to go deep into the tree to examine individual subtrees before moving on to the next subtree. This way, we can use a postorder traversal (left-right-root) to compute the sum and node count of each subtree.

While traversing, for each node, we calculate:

1. The total sum of node values (s) in the current subtree (includes the current node, its left subtree, and its right subtree).
2. The total number of nodes (n) in the current subtree.

With both these values at hand for each subtree, we then calculate the average by dividing s by n. This average is then compared to the current maximum average we've found (ans). If the current average is greater, we update ans with this new value.

The purpose of the nonlocal ans in the solution code is to allow the nested dfs function to modify the ans variable defined in the enclosing scope, which maintains the current maximum average.

As we recursively return the sum and node count for each subtree, we can easily compute the average for each parent subtree, inherently checking every subtree in the binary tree. The function ultimately returns the maximum average found during the traversal.

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

Solution Approach

The solution uses a recursive function to traverse the tree using depth-first search (DFS), specifically a postorder traversal pattern. In postorder traversal, we visit the left and right subtrees before processing the current node. This allows us to gather information from the subtrees before deciding on the parent node's information, which is essential for this problem as we need to compute subtrees' sums and node counts.

Here's a breakdown of the solution implementation:

1. Define a recursive function dfs that takes a node as an argument and returns a tuple (s, n) where:

   • s is the sum of all nodes within the current subtree.
   • n is the count of the nodes within the current subtree.

2. When the dfs function is called with None (i.e., the subtree is empty), it returns (0, 0), indicating that the sum and count are both 0 for an empty subtree.

3. The function calculates ls and ln by recursively calling dfs on the left child of the current node, and similarly rs and rn for the right child.

4. Once we have the sums (ls and rs) and counts (ln and rn) for both child subtrees, compute the sum s for the current node, which is the value of the current node plus ls plus rs.

5. Compute the count n for the current node as 1 (to count the current node) plus ln plus rn.

6. Use a nonlocal variable ans to keep track of the current maximum average. This variable is defined outside the dfs function so that it retains its value across different calls to dfs and isn't reinitialized in each recursive call.

7. Update ans with the maximum value between the current ans and the average for the current subtree (s / n).

8. After processing the root node, the dfs traversal ensures that each node's subtree has been considered. The ans variable will contain the maximum average.

9. The main function of the class Solution will initialize ans to 0 and call the dfs function on the binary tree's root.

10. Lastly, return the ans variable as the final result, which represents the maximum average value of a subtree of the binary tree.

This approach is efficient because it traverses each node of the tree exactly once, resulting in an O(n) time complexity, where n is the number of nodes in the binary tree.

The code for calculating sum and number of nodes in each subtree looks like this:

1if root is None:
2    return 0, 0
3ls, ln = dfs(root.left) # Postorder traversal - Left
4rs, rn = dfs(root.right) # Postorder traversal - Right
5s = root.val + ls + rs # Sum of current subtree
6n = 1 + ln + rn # Number of nodes in current subtree

And updating the maximum average found so far is as simple as:

1nonlocal ans
2ans = max(ans, s / n)

By carefully updating the ans variable during the traversal, we ensure that we have the maximum average value of any subtree by the end of the traversal.

Example Walkthrough

Let's walk through a simple example to illustrate the solution approach. Consider a binary tree:

1       5
2      / \
3     6   8
4    /   / \
5   1   7   3

We want to find the maximum average value of any subtree.

• Start by calling the dfs function on the root. We initialize ans = 0.

• At Node 5 - the root:

  1. First, call dfs on the left subtree (Node 6).
     • At Node 6:
       • Call dfs on the left subtree (Node 1), which returns (1, 1) - its own value and count since it is a leaf node.
       • Call dfs on the right subtree (None), which returns (0, 0).
       • Now, the sum at Node 6 is 6 (itself) + 1 (left subtree) = 7 and the count is 1 (itself) + 1 (left subtree) = 2.
       • Update the maximum average ans if the average here (7 / 2 = 3.5) is greater than the current ans.
  2. Then, call dfs on the right subtree (Node 8).
     • At Node 8:
       • Call dfs on the left subtree (Node 7), returning (7, 1).
       • Call dfs on the right subtree (Node 3), returning (3, 1).
       • Sum at Node 8 is 8 + 7 + 3 = 18 and count is 1 + 1 + 1 = 3.
       • Update the maximum average ans if the average here (18 / 3 = 6) is greater than the current ans.
  3. Now, the sum at Node 5 is 5 (itself) + 7 (left subtree) + 18 (right subtree) = 30 and the count is 1 (itself) + 2 (left subtree) + 3 (right subtree) = 6.
  4. Update the maximum average ans if the average here (30 / 6 = 5) is greater than the current ans.

• After traversing the entire tree, the ans variable holds the maximum average found, which is 6 in this case (the average value of the subtree rooted with Node 8).

This example illustrates the process of traversing the tree using postorder traversal, where we calculate the sum and count of the nodes for each subtree, updating the maximum average found (ans) along the way. The tree is only traversed once, making the solution efficient with O(n) complexity, where n is the number of nodes in the binary tree.

Solution Implementation

Time and Space Complexity

The given Python code defines a function maximumAverageSubtree that calculates the maximum average value of all subtrees in a binary tree. The dfs function, a helper function, is used to perform a depth-first search on the tree.

• Time Complexity:

The time complexity for this code is determined by the need to visit each node of the binary tree exactly once during the DFS traversal. Since every node is visited exactly once, and at every node we perform a constant amount of work (calculating sums and the number of nodes, and updating the maximum average), the time complexity is O(N), where N is the number of nodes in the binary tree.

• Space Complexity:

The space complexity is a bit more tricky to analyze due to the recursive nature of the DFS. In the worst case, the space complexity is defined by the height of the tree, which in the case of a skewed tree (where every node has only one child) could be O(N). However, for a balanced tree, the height (thus the maximum depth of the recursive call stack) would be O(log N). Thus, the space complexity of the code is O(H), where H is the height of the tree. In the average case for a reasonably balanced tree, this can be expressed as O(log N), but in the worst-case scenario (a completely unbalanced tree), the space complexity would be O(N).

To summarize:

• Time Complexity: O(N)
• Space Complexity: O(H), which would be O(log N) in a balanced tree and O(N) in the worst case.

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