Problem Description

You are given a binary tree with a root node. Your task is to remove exactly one edge from the tree, which will split it into two separate subtrees. After removing an edge, calculate the sum of all nodes in each of the two resulting subtrees, then multiply these two sums together to get their product.

Your goal is to find which edge to remove such that the product of the two subtree sums is maximized. Return this maximum product modulo 10^9 + 7.

Key points to understand:

• You must remove exactly one edge from the tree
• Removing an edge creates two separate subtrees
• Each subtree's sum is the total of all node values in that subtree
• You want to maximize the product: (sum of subtree 1) × (sum of subtree 2)
• The final answer should be returned modulo 10^9 + 7, but you need to find the maximum product first before applying the modulo operation

For example, if removing an edge creates two subtrees with sums of 6 and 9, the product would be 6 × 9 = 54. You need to try all possible edge removals and find which one gives the largest product.

The solution approach involves:

1. First calculating the total sum s of all nodes in the entire tree
2. Then traversing the tree again, and for each subtree with sum t, the other subtree would have sum s - t
3. The product for this split would be t × (s - t)
4. Tracking the maximum product across all possible splits
5. Returning the maximum product modulo 10^9 + 7

Flowchart Walkthrough

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

Is it a graph?

• Yes: A binary tree is a special type of graph where each node has at most two children and there are no cycles.

Is it a tree?

• Yes: The problem explicitly states we're working with a binary tree structure with a root node and parent-child relationships.

DFS

• Yes: We arrive at DFS (Depth-First Search) as the recommended approach.

Conclusion: The flowchart suggests using DFS for this tree problem.

Why DFS is the Right Choice

The DFS pattern is particularly well-suited for this problem because:

1. Tree Traversal: We need to visit every node in the tree to calculate subtree sums, which DFS handles naturally through recursive traversal.

2. Subtree Sum Calculation: DFS allows us to calculate the sum of each subtree in a bottom-up manner. As we return from recursive calls, we can aggregate the sums from child nodes.

3. Edge Removal Simulation: When we're at any node during DFS traversal, the sum of that node's subtree represents one part if we "cut" the edge connecting this node to its parent. The other part would be the total sum minus this subtree's sum.

4. Optimal Substructure: The problem exhibits optimal substructure - to find the sum of a subtree rooted at a node, we need the sums of its left and right subtrees plus the node's value itself.

The solution uses DFS twice:

• First DFS: Calculate the total sum of all nodes in the tree
• Second DFS: Traverse the tree again, and for each subtree, calculate the product if we remove the edge above it

This two-pass DFS approach efficiently explores all possible edge removals in O(n) time, where n is the number of nodes in the tree.

Intuition

The key insight is that when we remove any edge from a tree, we always get exactly two subtrees. If we know the total sum of all nodes in the original tree, and we know the sum of one subtree, we can immediately calculate the sum of the other subtree by subtraction.

Think about it this way: if the total sum is S and one subtree has sum t, then the other subtree must have sum S - t. The product would be t × (S - t).

This observation leads us to a clever approach: instead of actually removing edges and calculating sums for both resulting subtrees each time, we can:

1. Pre-calculate the total sum S of the entire tree
2. During a tree traversal, for each node, calculate the sum of the subtree rooted at that node
3. If we "cut" the edge above this node, one part would have sum t (the subtree), and the other would have sum S - t

Why does this work? When we're at a node during our traversal and calculate its subtree sum, we're essentially simulating what would happen if we cut the edge connecting this node to its parent. The subtree rooted at this node becomes one independent tree, and everything else becomes the other tree.

The beauty of using DFS here is that it naturally computes subtree sums in a bottom-up manner. As we recursively traverse the tree:

• We reach the leaves first (base case: sum = node value)
• As we backtrack, we accumulate sums: node.val + sum(left_subtree) + sum(right_subtree)
• Each recursive call returns the sum of its subtree, which represents a potential cut point

We need to check all possible cuts (all edges) to find the maximum product. Since each edge corresponds to a parent-child relationship, and we visit each node exactly once during DFS, we automatically consider all possible edge removals. The constraint t < s in the solution ensures we only consider valid cuts (we don't want to multiply by zero or negative values).

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

Solution Approach

The implementation uses a two-pass DFS strategy with clever optimization to avoid redundant calculations.

First Pass - Calculate Total Sum: The sum function performs a simple DFS to calculate the total sum of all nodes in the tree:

def sum(root: Optional[TreeNode]) -> int:
    if root is None:
        return 0
    return root.val + sum(root.left) + sum(root.right)

This gives us the total sum s which we'll use to calculate the complement of each subtree.

Second Pass - Find Maximum Product: The dfs function traverses the tree again, but this time it:

1. Calculates each subtree sum t using the same bottom-up approach
2. For each valid subtree (where t < s), calculates the product t × (s - t)
3. Tracks the maximum product found so far

def dfs(root: Optional[TreeNode]) -> int:
    if root is None:
        return 0
    t = root.val + dfs(root.left) + dfs(root.right)
    nonlocal ans, s
    if t < s:
        ans = max(ans, t * (s - t))
    return t

Key Implementation Details:

1. Why t < s check? This ensures we're considering a proper subtree, not the entire tree. When t = s, we're at the root and haven't removed any edge yet.

2. Use of nonlocal: The variables ans and s are declared as nonlocal to allow modification within the nested function. This is Python's way of handling closures when we need to update outer scope variables.

3. Modulo Operation: The modulo 10^9 + 7 is applied only at the final return, not during the calculation. This is crucial because we need to find the actual maximum product first, then reduce it modulo.

4. Time Complexity: O(n) where n is the number of nodes, as we visit each node exactly twice (once in each DFS pass).

5. Space Complexity: O(h) where h is the height of the tree, due to the recursive call stack.

The algorithm elegantly avoids the naive approach of actually removing each edge and recalculating sums from scratch, which would be O(n²). Instead, by pre-calculating the total and using the subtraction trick, we achieve linear time complexity.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Consider this binary tree:

        5
       / \
      3   2
     /
    1

Step 1: Calculate Total Sum First, we calculate the total sum of all nodes:

• Starting from root: 5 + (left subtree) + (right subtree)
• Left subtree of 5: 3 + 1 = 4
• Right subtree of 5: 2
• Total sum s = 5 + 4 + 2 = 11

Step 2: Traverse and Find Maximum Product Now we perform DFS again, calculating subtree sums and checking products:

1. Visit node 1 (leaf)

   • Subtree sum t = 1
   • Other part would have sum = 11 - 1 = 10
   • Product = 1 × 10 = 10
   • Update max product to 10

2. Visit node 3

   • Subtree sum t = 3 + 1 = 4
   • Other part would have sum = 11 - 4 = 7
   • Product = 4 × 7 = 28
   • Update max product to 28

3. Visit node 2 (leaf)

   • Subtree sum t = 2
   • Other part would have sum = 11 - 2 = 9
   • Product = 2 × 9 = 18
   • Max product remains 28

4. Visit node 5 (root)

   • Subtree sum t = 5 + 4 + 2 = 11
   • Since t = s, we skip this (no edge removed)

Result: The maximum product is 28, achieved by removing the edge between nodes 5 and 3, creating subtrees with sums 4 and 7.

Let's verify: If we remove the edge between 5 and 3:

• Subtree 1: Contains nodes {3, 1} with sum = 4
• Subtree 2: Contains nodes {5, 2} with sum = 7
• Product = 4 × 7 = 28 ✓

The final answer would be 28 % (10^9 + 7) = 28.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the number of nodes in the binary tree.

The algorithm consists of two tree traversals:

1. The sum() function traverses the entire tree once to calculate the total sum, visiting each node exactly once - O(n)
2. The dfs() function traverses the entire tree once more to calculate subtree sums and find the maximum product, visiting each node exactly once - O(n)

Total time complexity: O(n) + O(n) = O(n)

Space Complexity: O(h) where h is the height of the binary tree.

The space complexity is determined by the recursion call stack:

• Both sum() and dfs() functions use recursion that goes as deep as the height of the tree
• In the worst case (skewed tree), the height h = n, giving O(n) space complexity
• In the best case (balanced tree), the height h = log(n), giving O(log n) space complexity
• The additional variables (mod, s, ans) use O(1) space

Therefore, the space complexity is O(h) which ranges from O(log n) to O(n) depending on the tree structure.

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

Common Pitfalls

1. Applying Modulo During Calculation Instead of at the End

The Pitfall: A common mistake is to apply the modulo operation while calculating and comparing products:

# WRONG APPROACH
def find_max_product(node):
    if node is None:
        return 0
  
    subtree_sum = node.val + find_max_product(node.left) + find_max_product(node.right)
  
    if subtree_sum < total_sum:
        # Applying modulo here loses information!
        product = (subtree_sum * (total_sum - subtree_sum)) % MOD
        max_product = max(max_product, product)
  
    return subtree_sum

Why This Fails: When you apply modulo during the calculation, you lose the actual magnitude of the products. For example:

• Product A: 1000000014 % (10^9 + 7) = 7
• Product B: 1000000010 % (10^9 + 7) = 3
• After modulo, it appears B < A, but actually A > B

The Solution: Always find the actual maximum first, then apply modulo only at the final return:

# CORRECT APPROACH
max_product = max(max_product, subtree_sum * (total_sum - subtree_sum))
# ... continue finding maximum
return max_product % MOD  # Apply modulo only at the end

2. Integer Overflow in Other Languages

The Pitfall: In languages like Java or C++, the product subtree_sum * (total_sum - subtree_sum) can overflow even with 64-bit integers if the tree has large values.

The Solution: Use appropriate data types or cast to larger types before multiplication:

// Java example
long product = (long)subtreeSum * (totalSum - subtreeSum);

// C++ example
long long product = static_cast<long long>(subtreeSum) * (totalSum - subtreeSum);

3. Forgetting the Edge Case Check (t < s)

The Pitfall: Omitting the if subtree_sum < total_sum check would incorrectly consider the entire tree as a "subtree":

# WRONG: Missing the check
def find_max_product(node):
    if node is None:
        return 0
  
    subtree_sum = node.val + find_max_product(node.left) + find_max_product(node.right)
  
    # This would calculate total_sum * 0 when at root!
    max_product = max(max_product, subtree_sum * (total_sum - subtree_sum))
  
    return subtree_sum

Why This Fails: When we reach the root node after processing all children, subtree_sum equals total_sum. This would give us a product of total_sum * 0 = 0, which might incorrectly be chosen as the maximum if all valid products are negative (though this is impossible with positive node values in this problem).

The Solution: Always include the check to ensure we're considering a proper subtree:

if subtree_sum < total_sum:  # Only consider proper subtrees
    max_product = max(max_product, subtree_sum * (total_sum - subtree_sum))

4. Not Handling the Nonlocal Variables Correctly

The Pitfall: Forgetting to declare variables as nonlocal or trying to use global incorrectly:

# WRONG: Creates local variables instead of modifying outer scope
def find_max_product(node):
    if node is None:
        return 0
  
    subtree_sum = node.val + find_max_product(node.left) + find_max_product(node.right)
  
    # This creates new local variables, doesn't update the outer ones!
    max_product = max(max_product, subtree_sum * (total_sum - subtree_sum))
  
    return subtree_sum

The Solution: Properly declare nonlocal variables:

def find_max_product(node):
    nonlocal max_product, total_sum  # Declare at the beginning
    # ... rest of the function