Solution Explanation for Diameter of N-Ary Tree

This problem asks for the diameter of an N-ary tree, which is the length of the longest path between any two nodes. The path doesn't necessarily pass through the root. Two efficient solutions are presented below.

Solution 1: Depth-First Search (DFS)

This solution uses a depth-first search approach to efficiently find the diameter. The key idea is that the diameter either passes through the root or is entirely within one of the subtrees.

Algorithm:

1. dfs(root) function: This recursive function calculates the maximum depth of a subtree rooted at root and updates the global maximum diameter ans.

2. Finding the two longest paths: For each child of the current node, the dfs function is recursively called to find the depth of its subtree. The two largest depths (m1 and m2) are tracked.

3. Updating the diameter: The current diameter ans is updated to the maximum of its current value and the sum of m1 and m2 (representing the longest path through the current node).

4. Returning the maximum depth: The function returns 1 + m1, representing the maximum depth of the subtree rooted at the current node (including the current node itself).

Time Complexity: O(N), where N is the number of nodes in the tree. Each node is visited exactly once during the DFS traversal.

Space Complexity: O(H), where H is the height of the tree. This is due to the recursion stack used during the DFS traversal. In the worst case, the height can be equal to N (a skewed tree).

Solution 2: Breadth-First Search (BFS) or Graph Approach

This approach treats the N-ary tree as a graph. It builds an adjacency list representing the connections between nodes, then performs two BFS traversals to find the diameter.

Algorithm:

1. build(root) function: This function constructs the adjacency list g. For each node, it adds its children and parent to the adjacency list.

2. dfs(u, t) function: This function performs a breadth-first search starting from node u. It keeps track of the distance (t) from the starting node.

3. Finding the farthest node: The first DFS finds the farthest node (next) from the root.

4. Finding the diameter: The second DFS starts from next and finds the farthest node from it. The distance to this farthest node is the diameter.

Time Complexity: O(N + E), where N is the number of nodes and E is the number of edges. Building the adjacency list takes O(N + E) time. Each BFS traversal takes O(N + E) time. In N-ary tree E <= N*degree, thus is roughly O(N)

Space Complexity: O(N) to store the adjacency list g.

Code Examples (Python, Java, C++, Go): The code examples provided previously accurately reflect these algorithms. Note that the Python solution uses defaultdict(set) for concise adjacency list representation and the other languages use HashMap or similar data structures.

Choosing the Right Solution:

Solution 1 (DFS) is generally preferred due to its simplicity and slightly better space complexity in many cases. Solution 2 (BFS/Graph) might be advantageous if you need the adjacency list for other operations on the tree. The time complexity of both approaches is effectively linear, making either a valid choice for most scenarios.