Problem Description

You are given a tree with n nodes numbered from 0 to n - 1, rooted at node 0. The tree structure is represented by a 2D integer array edges, where edges[i] = [ui, vi] indicates there is an edge between nodes ui and vi.

Each node has a color assigned to it, given in a 0-indexed integer array colors of size n, where colors[i] represents the color of node i.

Your task is to find a node v such that every node in the subtree rooted at v (including v itself) has the same color. Among all such valid subtrees, you need to return the size (number of nodes) of the largest one.

In other words, you need to:

1. Identify all subtrees where all nodes share the same color
2. Return the maximum number of nodes among these valid subtrees

For example, if a subtree rooted at node v contains nodes {v, a, b, c} and all four nodes have the same color, this would be a valid subtree of size 4. You need to find the largest such valid subtree in the entire tree.

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 a tree structure with nodes and edges. A tree is a special type of graph (connected and acyclic).

Is it a tree?

• Yes: The problem clearly states we have a tree with n nodes numbered from 0 to n-1, rooted at node 0. The edges array represents the tree structure where edges[i] = [ui, vi] indicates connections between nodes.

DFS

• According to the flowchart, when we have a tree structure, we should use DFS (Depth-First Search).

Why DFS is appropriate for this problem:

1. We need to explore subtrees - DFS naturally traverses subtrees completely before moving to siblings
2. We need to check if all nodes in a subtree have the same color - this requires visiting all descendants of a node
3. We need to calculate subtree sizes - DFS allows us to aggregate information from children nodes as we backtrack
4. The problem requires examining every possible subtree rooted at each node - DFS systematically visits each node and can treat it as a potential subtree root

Conclusion: The flowchart correctly leads us to use DFS for this tree-based problem. DFS allows us to traverse each subtree, check color consistency within subtrees, calculate subtree sizes, and track the maximum valid subtree size - all in a single traversal of the tree.

Intuition

When dealing with tree problems where we need to examine properties of subtrees, DFS is a natural choice because it completely explores one branch before moving to another. This problem asks us to find the largest subtree where all nodes share the same color.

The key insight is that we can check each node as a potential root of a valid same-color subtree. For any node to be the root of a valid subtree:

1. All its descendants must have the same color as itself
2. Each of its child subtrees must also be valid (all same color)

This suggests a bottom-up approach: start from the leaves and work our way up. Why? Because:

• Leaf nodes trivially form valid single-node subtrees
• For any internal node, we can determine if it forms a valid subtree by checking if all its children have the same color as itself AND if each child's subtree is also valid
• As we traverse back up, we can accumulate the size of each subtree

The DFS traversal naturally gives us this bottom-up processing through recursion. When we visit a node:

1. First, we recursively process all its children (going deep)
2. Then, we check if the current node and all its children have matching colors
3. If yes, we've found a valid subtree - update our answer if it's the largest so far
4. Return whether this subtree is valid to help parent nodes make their decision

The beauty of this approach is that we only need one pass through the tree. Each node is visited once, and we make our decision about each potential subtree based on information gathered from its children. The recursive nature of DFS perfectly matches the recursive structure of the tree, making it elegant to check color consistency while simultaneously calculating subtree sizes.

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

Solution Approach

The implementation follows the DFS approach we identified, with several key components:

Data Structure Setup:

• Build an adjacency list g where g[a] stores all nodes connected to node a. This allows efficient traversal of the tree.
• Initialize a size array where size[a] tracks the number of nodes in the subtree rooted at node a. Initially, each node has size 1 (itself).

DFS Function Design: The core logic is in the dfs(a, fa) function, where:

• a is the current node being processed
• fa is the parent of node a (to avoid revisiting in an undirected tree)
• Returns a boolean indicating whether the subtree rooted at a has all nodes of the same color

DFS Processing Steps:

1. Initialize tracking variable: Start with ok = True, assuming the subtree is valid until proven otherwise.

2. Process all children: For each neighbor b of node a:

   • Skip if b is the parent fa (avoid going backward in the tree)
   • Recursively call dfs(b, a) to check if child b's subtree is valid
   • Update ok using the logical AND of three conditions:
     • Current ok status
     • colors[a] == colors[b] (parent and child have same color)
     • t (the child's subtree is valid)
   • Accumulate subtree size: size[a] += size[b]

3. Update answer if valid: If ok remains True after checking all children, we've found a valid same-color subtree. Update the global answer: ans = max(ans, size[a])

4. Return validity: Return ok to inform the parent whether this subtree is valid.

Algorithm Execution:

• Start DFS from the root node: dfs(0, -1) where -1 indicates no parent
• The DFS will explore the entire tree, checking every possible subtree
• The maximum valid subtree size encountered is stored in ans

Time Complexity: O(n) where n is the number of nodes, as each node is visited exactly once.

Space Complexity: O(n) for the adjacency list, size array, and recursion stack in the worst case (skewed tree).

Example Walkthrough

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

Consider a tree with 5 nodes and the following structure:

    0 (color: 2)
   / \
  1   2
(color: 2) (color: 2)
       / \
      3   4
   (color: 2) (color: 1)

Given:

• edges = [[0,1], [0,2], [2,3], [2,4]]
• colors = [2, 2, 2, 2, 1]

Step-by-step DFS execution:

Initial Setup:

• Build adjacency list: g = {0: [1,2], 1: [0], 2: [0,3,4], 3: [2], 4: [2]}
• Initialize size = [1, 1, 1, 1, 1] (each node starts with size 1)
• ans = 0 (tracks maximum valid subtree size)

DFS Traversal from root (node 0):

1. Visit node 0 (parent = -1)

   • Process child 1: Call dfs(1, 0)

     • Node 1 has no children (except parent 0), so ok = True
     • Node 1 forms valid subtree of size 1
     • Update ans = max(0, 1) = 1
     • Return True to node 0

   • Back at node 0: Check colors[0] == colors[1] → 2 == 2 ✓

   • Update size[0] = 1 + 1 = 2

   • Process child 2: Call dfs(2, 0)

     • Process child 3: Call dfs(3, 2)

       • Node 3 has no children, forms valid subtree of size 1
       • Update ans = 1 (no change)
       • Return True to node 2

     • Check colors[2] == colors[3] → 2 == 2 ✓

     • Update size[2] = 1 + 1 = 2

     • Process child 4: Call dfs(4, 2)

       • Node 4 has no children, forms valid subtree of size 1
       • Update ans = 1 (no change)
       • Return True to node 2

     • Check colors[2] == colors[4] → 2 == 1 ✗

     • ok becomes False for node 2's subtree

     • size[2] = 2 + 1 = 3 (still accumulate size)

     • Since ok = False, don't update ans

     • Return False to node 0

   • Back at node 0: Child 2's subtree is not valid (False)

   • ok becomes False for node 0's subtree

   • size[0] = 2 + 3 = 5

   • Since ok = False, don't update ans

Valid subtrees found:

• Node 1: size 1, all nodes have color 2 ✓
• Node 3: size 1, all nodes have color 2 ✓
• Node 4: size 1, all nodes have color 1 ✓
• Node 2's subtree: size 3, but mixed colors (2 and 1) ✗
• Node 0's subtree: size 5, but mixed colors (2 and 1) ✗

Result: The maximum valid subtree size is 1.

If we changed node 4's color to 2, then:

• Node 2's subtree would be valid with size 3 (nodes 2, 3, 4 all color 2)
• Node 0's subtree would be valid with size 5 (all nodes color 2)
• The answer would be 5

This example demonstrates how the algorithm:

1. Traverses the tree depth-first
2. Checks color consistency between parent and children
3. Propagates validity information upward
4. Tracks the maximum size of valid subtrees encountered

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. For each node, the algorithm:

• Iterates through all its adjacent nodes (children in the tree)
• Performs constant-time operations (comparisons and updates)

Since the input represents a tree with n nodes and n-1 edges, the total number of edges traversed is 2(n-1) (each edge is traversed twice, once from each direction). Therefore, the overall time complexity is O(n).

Space Complexity: O(n)

The space complexity consists of:

• The adjacency list g: O(n) space to store all nodes and their connections
• The size array: O(n) space to store the size of each subtree
• The recursion call stack: O(h) where h is the height of the tree, which in the worst case (skewed tree) can be O(n)

The dominant factor is O(n), making the overall space complexity O(n).

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

Common Pitfalls

1. Incorrectly Handling Tree Structure Assumptions

Pitfall: The code assumes the tree is always rooted at node 0, but the problem statement might not guarantee this. Additionally, the code assumes nodes are numbered from 0 to n-1 consecutively, which could fail if there are gaps in node numbering.

Solution: Validate the tree structure and handle edge cases:

def maximumSubtreeSize(self, edges: List[List[int]], colors: List[int]) -> int:
    if not edges:
        return 1  # Single node tree
  
    # Find all unique nodes to handle non-consecutive numbering
    nodes = set()
    for u, v in edges:
        nodes.add(u)
        nodes.add(v)
  
    num_nodes = len(nodes)
    # Verify we have the right number of colors
    if len(colors) != num_nodes:
        raise ValueError("Mismatch between number of nodes and colors")
  
    # Continue with adjacency list building...

2. Short-Circuit Evaluation Breaking Subtree Size Calculation

Pitfall: Using short-circuit evaluation incorrectly when checking validity could prevent proper subtree size accumulation:

# WRONG: This might skip size calculation if is_valid_subtree becomes False
if is_valid_subtree:
    child_is_valid = dfs(neighbor, current_node)
    is_valid_subtree = colors[current_node] == colors[neighbor] and child_is_valid
    subtree_sizes[current_node] += subtree_sizes[neighbor]

Solution: Always perform DFS and size calculation, then update validity:

# CORRECT: Always calculate sizes regardless of validity
child_is_valid = dfs(neighbor, current_node)
subtree_sizes[current_node] += subtree_sizes[neighbor]  # Always accumulate size
is_valid_subtree = (is_valid_subtree and 
                   colors[current_node] == colors[neighbor] and 
                   child_is_valid)

3. Not Considering Disconnected Components

Pitfall: The code assumes a connected tree, but if the input has disconnected components or is a forest, only the component containing node 0 will be processed.

Solution: Check for connectivity or process all components:

def maximumSubtreeSize(self, edges: List[List[int]], colors: List[int]) -> int:
    num_nodes = len(colors)  # Use colors length as node count
    if num_nodes == 0:
        return 0
    if num_nodes == 1:
        return 1
  
    # Check if we have a valid tree (n-1 edges for n nodes)
    if len(edges) != num_nodes - 1:
        # Handle forest case - process each component separately
        visited = [False] * num_nodes
        max_subtree_size = 0
      
        for start_node in range(num_nodes):
            if not visited[start_node]:
                component_max = self.process_component(start_node, visited, ...)
                max_subtree_size = max(max_subtree_size, component_max)
      
        return max_subtree_size

4. Integer Overflow in Large Trees

Pitfall: While Python handles arbitrary precision integers, in other languages or when porting this solution, integer overflow could occur when accumulating subtree sizes.

Solution: Use appropriate data types and add bounds checking:

# Add validation for tree size
if num_nodes > 10**5:  # Typical constraint
    raise ValueError("Tree too large")

# In languages like Java/C++, use long/long long for size arrays

5. Forgetting to Handle Single Node Trees

Pitfall: When edges is empty (single node tree), the code might fail or return 0 instead of 1.

Solution: Add explicit handling for edge cases:

def maximumSubtreeSize(self, edges: List[List[int]], colors: List[int]) -> int:
    # Handle single node case
    if not edges and len(colors) == 1:
        return 1
  
    if not edges or not colors:
        return 0
  
    # Rest of the implementation...