Tree Diameter

class TreeAnalysis {
    private List<List<Integer>> adjacencyList;
    private Integer maxDiameter = 0;
    
    public int calculateTreeDiameter(int[][] edges) {
        // Initialize the adjacency list and visited array.
        this.adjacencyList = new ArrayList<List<Integer>>();
        boolean[] seen = new boolean[edges.length + 1];
        for (int i = 0; i < edges.length + 1; ++i) {
            this.adjacencyList.add(new ArrayList<Integer>());
            seen[i] = false;
        }
        for (int[] edge : edges) {
            Integer node1 = edge[0], node2 = edge[1];
            this.adjacencyList.get(node1).add(node2);
            this.adjacencyList.get(node2).add(node1);
        }
    
        depthFirstSearch(0, seen);
    
        return this.maxDiameter;
    }
    
    private int depthFirstSearch(int node, boolean[] seen) {
            
        Integer greatest = 0, secondGreatest = 0;
    
        seen[node] = true;
        for (Integer neighbor : adjacencyList.get(node)) {
            int dist = 0;
            if (!seen[neighbor])
                dist = 1 + this.depthFirstSearch(neighbor, seen);
    
            if (dist > greatest) {
                secondGreatest = greatest;
                greatest = dist;
            } else if (dist > secondGreatest) {
                secondGreatest = dist;
            }
        }
            
        // Update maximum diameter at this node
        this.maxDiameter = Math.max(this.maxDiameter, greatest + secondGreatest);
    
        return greatest;
    }
}

The Tree Diameter problem requires finding the longest path between any two nodes in a tree. Here's a brief summary of how the given Java solution approaches this:

• The code defines a class TreeAnalysis containing methods to calculate the tree's diameter based on an adjacency list representation of the tree.
• The calculateTreeDiameter method initializes the adjacency list from the provided edge list of the tree and also maintains a seen array to keep track of visited nodes.
• The method utilizes a helper function depthFirstSearch that recursively explores each node, calculating the greatest and the second greatest distances to its children. This helps in determining the path length through the current node by summing the two longest child paths.
• During each recursion, the function updates a class variable maxDiameter which keeps track of the maximum diameter found during the traversal.

Critical operations in this solution include:

• Initializing and filling an adjacency list from the input array of edges.
• Using a Depth First Search (DFS) approach to explore the tree and calculate distances.
• Tracking the two longest path lengths from any node to determine potential diameters and updating a global maximum accordingly.

Ensure the input tree is not empty or null to avoid runtime errors, and the solution efficiently determines the tree diameter with optimal tree traversal and path tracking techniques.

class Solution:
    def calculateTreeDiameter(self, connections: List[List[int]]) -> int:
    
        # Construct the adjacency list for the tree
        adjacency = [set() for _ in range(len(connections) + 1)]
        for connection in connections:
            p1, p2 = connection
            adjacency[p1].add(p2)
            adjacency[p2].add(p1)
    
        maxDiameter = 0
    
        def dfs(node, seen):
            max1, max2 = 0, 0  # initialize the two maximum distances
    
            seen[node] = True
    
            for neighbor in adjacency[node]:
                if not seen[neighbor]:
                    distance = 1 + dfs(neighbor, seen)
    
                    if distance > max1:
                        max1, max2 = distance, max1
                    elif distance > max2:
                        max2 = distance
    
            # update the maximum diameter found so far
            nonlocal maxDiameter
            maxDiameter = max(maxDiameter, max1 + max2)
    
            return max1
    
        visited = [False] * len(adjacency)
        dfs(0, visited)
    
        return maxDiameter

The given Python code provides a solution to find the diameter of a tree, where the diameter is defined as the longest path between any two nodes in the tree. Here's an outline of the solution approach:

1. The function calculateTreeDiameter initializes with a set of connections that define the tree edges.
2. An adjacency list, adjacency, is constructed to represent the tree structure. Each node points to a set of its connected neighbors.
3. The dfs function is implemented to traverse the tree depth-first. This function helps in exploring the maximum distances from a given node to its farthest children. It uses three key variables:
   • max1 for the maximum distance
   • max2 for the second maximum distance
   • seen list tracks the visited nodes to avoid cycles, which is crucial since the tree does not naturally contain any cycles.
4. Each call of dfs updates the maxDiameter which is a nonlocal variable, ensuring that changes persist outside the function's current scope.
5. The path of maximum length is continuously updated if the sum of the two largest distances found (max1 + max2) exceeds the current maxDiameter.
6. The solution process starts by initializing the visited array to keep track of visited nodes and prevent revisiting. The DFS is initiated from node 0 — often presumed to be the root in such tree representations.
7. After the DFS traversal is complete, maxDiameter holds the length of the longest path in the tree, and this value is returned by the function.

This code effectively utilizes recursion, graph representation, and depth-first search techniques to solve the problem efficiently. Be mindful that in tree structures, avoiding the revisiting of nodes is critical, hence the use of a visited array or set is standard practice. The solution shown is both efficient in terms of time complexity (primarily O(n) where n is the number of nodes, due to the DFS traversal of the tree) and space complexity, primarily for storing the adjacency list and the visited nodes.