Problem Description

You have an undirected tree with n nodes labeled from 0 to n - 1. The tree structure is given by an array edges where each edges[i] = [ai, bi] represents an edge between nodes ai and bi. Each node has a price value given in the array price, where price[i] is the price of node i.

The key concepts to understand:

1. Path Price Sum: For any path in the tree, the price sum is the total of all node prices along that path.

2. Rooting the Tree: You can choose any node as the root. Once a root is chosen, all paths are considered as starting from that root and going downward through the tree.

3. Cost of a Root: After choosing a root node, the cost is calculated as the difference between:

   • The maximum price sum among all paths starting from the root
   • The minimum price sum among all paths starting from the root

4. Objective: Find the maximum possible cost across all possible choices of root nodes.

For example, if you choose node x as root and the paths from x have price sums of [10, 15, 8, 20], then:

• Maximum price sum = 20
• Minimum price sum = 8
• Cost for this root = 20 - 8 = 12

You need to try all possible root nodes and return the highest cost achievable.

The solution uses DFS (Depth First Search) to explore the tree. For each node, it tracks two values:

• a: Maximum path sum including the current node
• b: Maximum path sum excluding the current node (essentially the path ends at a child)

By maintaining these values and combining paths from different subtrees, the algorithm efficiently computes the maximum cost without explicitly trying every root.

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 explicitly mentions an undirected tree with nodes and edges. A tree is a special type of graph (connected and acyclic).

Is it a tree?

• Yes: The problem states "There exists an undirected and initially unrooted tree with n nodes" and provides edges that form a tree structure.

DFS

• Conclusion: Since we have a tree structure, the flowchart directly leads us to use DFS (Depth-First Search).

The flowchart confirms that DFS is the appropriate algorithm for this problem. This makes sense because:

1. Tree Traversal: We need to explore all possible paths from each potential root node, which is a natural fit for DFS on trees.

2. Path Calculations: DFS allows us to recursively calculate path sums from leaves up to the root, maintaining the maximum and minimum values as we traverse.

3. Subtree Information: For each node, we need to gather information from its subtrees (children), which DFS handles efficiently through its recursive nature.

4. Re-rooting Technique: The solution uses DFS to compute results for all possible roots in a single traversal by tracking two values at each node:

   • Maximum path sum including the current node
   • Maximum path sum excluding the current node

This allows us to efficiently calculate the maximum difference between path sums without explicitly re-rooting the tree multiple times.

Conclusion: The flowchart correctly identifies DFS as the optimal approach for this tree-based path sum problem.

Intuition

The key insight is that we want to find the maximum difference between two paths starting from the same root. The minimum path sum from any root is always just the root itself (a path of length 0), so we're essentially looking for the maximum path sum from each potential root.

But here's the clever part: instead of trying every node as root separately (which would be inefficient), we can solve this in a single DFS traversal by thinking about paths differently.

For any node in the tree, a path passing through it can be viewed as:

1. A path coming from one subtree, passing through the node, and potentially extending into another subtree
2. The combination of the node with paths from its children

This leads us to track two important values for each node:

• a: The maximum path sum that includes the current node (the path starts from somewhere below and goes up to include this node)
• b: The maximum path sum that excludes the current node (the path starts from somewhere below but doesn't include this node's price)

Why these two values? Because when we're at a node with multiple children:

• We can combine a path that includes prices from one child (a from child 1) with a path that excludes the current node from another child (b from child 2)
• Or vice versa (b from child 1 with a from child 2)

This combination gives us the maximum "spread" or difference we're looking for. The path including the node represents our maximum sum, while implicitly the minimum is just the node itself.

The beauty of this approach is that as we traverse up from the leaves, we're constantly computing the best possible paths and their differences at each node, effectively considering all possible root choices in a single pass.

For example, if we're at node i with children, and child j returns (c, d):

• c represents the best path in child j's subtree that includes j
• d represents the best path in child j's subtree that excludes j

We can then combine these with paths from other children to find the maximum difference at node i, which could be a candidate for our final answer.

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

Solution Approach

The implementation uses DFS with a clever state tracking mechanism. Let's walk through the key components:

Data Structure Setup:

• Build an adjacency list g using defaultdict(list) to represent the tree
• Since the tree is undirected, add edges in both directions

DFS Function Design:

The dfs(i, fa) function processes node i with parent fa and returns two values:

• a: Maximum path sum that includes node i
• b: Maximum path sum that excludes node i

Base Case: For a leaf node (no children except parent):

• a = price[i] (path is just the node itself)
• b = 0 (empty path, excluding the node)

Recursive Case: For each child j of node i:

1. Recursively call dfs(j, i) to get (c, d) where:

   • c: Maximum path including child j
   • d: Maximum path excluding child j

2. Update the global answer by considering two scenarios:

   • a + d: Current best path including i combined with path excluding child j
   • b + c: Current best path excluding i combined with path including child j

   These combinations represent complete paths through node i that maximize the difference.

3. Update local values:

   • a = max(a, price[i] + c): Best path including i is either just i itself or i plus the best path from a child
   • b = max(b, price[i] + d): Best path "excluding" i actually means the path ends at i but came from below, so we add price[i] to the child's excluded path

Key Insight in the Update Logic:

When we compute ans = max(ans, a + d, b + c), we're finding paths that:

• Start from one subtree
• Pass through node i
• End at node i or continue to another subtree

The difference between including/excluding helps us track whether we're counting the node's price once (as an intermediate node) or as an endpoint.

Final Answer: Start DFS from any node (here node 0) with parent -1, and return the maximum difference found across all nodes during the traversal.

Example Walkthrough

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

Consider a tree with 4 nodes and prices:

• Nodes: 0, 1, 2, 3
• Edges: [[0,1], [1,2], [1,3]]
• Prices: [2, 3, 5, 4]

The tree structure looks like:

      0 (price=2)
      |
      1 (price=3)
     / \
    2   3
  (5)   (4)

Step 1: Build adjacency list

g[0] = [1]
g[1] = [0, 2, 3]
g[2] = [1]
g[3] = [1]

Step 2: Start DFS from node 0

Call dfs(0, -1):

• Node 0 has one child: node 1
• Need to process dfs(1, 0) first

Step 3: Process dfs(1, 0)

Node 1 has children 2 and 3 (excluding parent 0).

First, process dfs(2, 1):

• Node 2 is a leaf (no children except parent 1)
• Returns (a=5, b=0)
• Meaning: max path including node 2 is 5, excluding is 0

Next, process dfs(3, 1):

• Node 3 is a leaf
• Returns (a=4, b=0)

Now at node 1:

• Initialize a = price[1] = 3, b = 0

Process child 2 with (c=5, d=0):

• Update answer: max(ans, a+d, b+c) = max(0, 3+0, 0+5) = 5
• Update a = max(3, 3+5) = 8 (path: 2→1)
• Update b = max(0, 3+0) = 3

Process child 3 with (c=4, d=0):

• Update answer: max(ans, a+d, b+c) = max(5, 8+0, 3+4) = 8
• Update a = max(8, 3+4) = 8 (still path: 2→1)
• Update b = max(3, 3+0) = 3

Node 1 returns (a=8, b=3) to node 0.

Step 4: Complete dfs(0, -1)

At node 0:

• Initialize a = price[0] = 2, b = 0
• Process child 1 with (c=8, d=3):
  • Update answer: max(ans, a+d, b+c) = max(8, 2+3, 0+8) = 8
  • Update a = max(2, 2+8) = 10 (path: 2→1→0)
  • Update b = max(0, 2+3) = 5

Returns (a=10, b=5)

Final Answer: 8

The maximum cost is 8, which represents:

• Choosing node 1 as root
• Maximum path sum: 8 (path from node 2 to node 1)
• Minimum path sum: 0 (empty path, just considering paths going down from root)
• Cost = 8 - 0 = 8

The algorithm found this optimal value when processing node 1, considering the path from node 2 (price 5) through node 1 (price 3), giving us the maximum difference of 8.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

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

• Iterates through all adjacent nodes (children in the tree)
• Performs constant time operations for updating variables a, b, and ans

Since this is a tree with n nodes and n-1 edges, the total number of edges traversed is O(n). Therefore, the overall time complexity is O(n).

Space Complexity: O(n)

The space complexity consists of:

• Graph representation: The adjacency list g stores all edges. Since there are n-1 edges in a tree and each edge is stored twice (bidirectional), this requires O(n) space.
• Recursion call stack: In the worst case (when the tree is a linear chain), the DFS recursion can go up to depth n, requiring O(n) stack space.
• Local variables: The variables a, b, c, d, and ans use O(1) space at each recursion level.

The dominant factors are the graph storage and recursion stack, both being O(n), resulting in an overall space complexity of O(n).

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

Common Pitfalls

Pitfall 1: Misunderstanding the Return Values

The Problem: A common mistake is misinterpreting what max_path_without_node represents. Despite its name suggesting "without node", it actually includes the current node's price (price[node] + child_without). This confuses many developers who expect it to completely exclude the current node.

Why It Happens: The variable naming is counterintuitive. When we say "without node", we're actually referring to paths that don't start from this node but pass through it as an intermediate point.

The Fix: Rename variables to better reflect their actual meaning:

def dfs(node: int, parent: int) -> tuple[int, int]:
    # Better naming
    max_path_starting_here = price[node]  # Path starts from this node
    max_path_through_here = 0  # Path passes through but doesn't start here

Pitfall 2: Incorrect Answer Updates

The Problem: Developers often update the answer incorrectly by trying combinations like:

max_result = max(max_result, child_with + child_without)  # Wrong!

This doesn't properly account for the current node's contribution to the path.

Why It Happens: The logic for combining paths from different subtrees through the current node is subtle. We need to ensure we're creating valid paths that include the current node exactly once.

The Fix: Always combine one path that includes the current node with one that doesn't:

# Correct: Combine paths properly through current node
max_result = max(max_result, 
                max_path_with_node + child_without,  # Path from one side including current
                max_path_without_node + child_with)  # Path from other side including current

Pitfall 3: Edge Case with Single Node

The Problem: When the tree has only one node, the DFS never updates max_result because there are no children to process, leading to a return value of 0 instead of the expected 0 (since max path = min path = node price).

Why It Happens: The answer is only updated when processing children. With no edges, the loop never executes.

The Fix: Handle the single node case explicitly:

def maxOutput(self, n: int, edges: List[List[int]], price: List[int]) -> int:
    if n == 1:
        return 0  # Single node: max_path = min_path = price[0], difference = 0
  
    # Rest of the implementation...

Pitfall 4: Forgetting to Use Nonlocal

The Problem: Forgetting to declare nonlocal max_result inside the DFS function causes either an error or creates a local variable that shadows the outer one:

def dfs(node, parent):
    max_result = max(max_result, ...)  # UnboundLocalError without nonlocal

Why It Happens: Python's scoping rules require explicit declaration when modifying variables from outer scopes.

The Fix: Always declare nonlocal before modifying:

def dfs(node, parent):
    nonlocal max_result  # Declare before use
    # ... rest of function

Complete Corrected Solution:

class Solution:
    def maxOutput(self, n: int, edges: List[List[int]], price: List[int]) -> int:
        # Handle edge case
        if n == 1:
            return 0
          
        from collections import defaultdict
      
        def dfs(node: int, parent: int) -> tuple[int, int]:
            # Clearer variable names
            path_from_here = price[node]  # Max path starting from this node
            path_through_here = 0         # Max path passing through (not starting from) this node
          
            for child in adjacency_list[node]:
                if child != parent:
                    child_from, child_through = dfs(child, node)
                  
                    # Update answer with valid path combinations
                    nonlocal max_cost
                    max_cost = max(max_cost,
                                  path_from_here + child_through,
                                  path_through_here + child_from)
                  
                    # Update local maximums
                    path_from_here = max(path_from_here, price[node] + child_from)
                    path_through_here = max(path_through_here, price[node] + child_through)
          
            return path_from_here, path_through_here
      
        adjacency_list = defaultdict(list)
        for a, b in edges:
            adjacency_list[a].append(b)
            adjacency_list[b].append(a)
      
        max_cost = 0
        dfs(0, -1)
        return max_cost