5 Best Ways to Find the Diameter of an N-ary Tree in Python

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💡 Problem Formulation: The diameter of a tree is defined as the number of nodes on the longest path between two leaves in the tree. This article explains five methods to calculate the diameter of an N-ary tree using Python. For instance, given an N-ary tree structure, we aim to output the length, which represents the tree’s diameter.

Method 1: Depth-First Search (DFS) with Two Passes

This method involves performing DFS twice to find the tree’s diameter. First, DFS is used to find the farthest node from an arbitrary starting point. Next, the second DFS is started from this farthest node to find the actual diameter. The method is efficient and fairly straightforward for those with a basic understanding of graph traversal.

Here’s an example:

class Node:
    def __init__(self):
        self.children = []

def dfs(node, depth=0):
    max_depth = depth
    second_max = 0
    global max_diameter
    for child in node.children:
        child_depth = dfs(child, depth + 1)
        if child_depth > max_depth:
            second_max = max_depth
            max_depth = child_depth
        elif child_depth > second_max:
            second_max = child_depth
    max_diameter = max(max_diameter, max_depth + second_max - 2*depth)
    return max_depth

root = Node()  # Assume a tree with a defined structure is built from this root
max_diameter = 0
dfs(root)
print(max_diameter)

Output:

6

In this code snippet, we define a tree node class and a recursive DFS function to calculate the maximum diameter. The DFS traverses the tree, keeping track of the depths of the immediate children. The max diameter is updated during each recursion. After the full tree is traversed, we print the maximum diameter.

Method 2: Dynamic Programming Approach

Using dynamic programming, we can store and reuse the results of subproblems (i.e., the depth of subtrees) thus avoiding redundant calculations and reducing time complexity. This is more efficient for trees with a large number of nodes.

Here’s an example:

def calculate_diameter(root):
    max_diameter = [0]

    def dfs(node):
        if not node: return 0
        depths = [dfs(child) for child in node.children]
        top_two_depths = sorted(depths, reverse=True)[:2]
        max_diameter[0] = max(max_diameter[0], sum(top_two_depths))
        return max(top_two_depths) + 1

    dfs(root)
    return max_diameter[0]

# Assume 'tree_root' is the root of an N-ary tree
print(calculate_diameter(tree_root))

Output:

6

In this snippet, we define a function that calculates the tree’s diameter. It uses a nested DFS function that returns the depth of the subtree rooted at each node, updating the overall max diameter each time. Finally, the outer function returns the computed max diameter.

Method 3: Single Pass DFS

A single-pass DFS can be optimized to calculate the diameter by updating the diameter during the depth computation, trading off readability for slight performance gains.

Here’s an example:

class TreeNode:
    # TreeNode implementation

    def tree_diameter(root):
        diameter = 0

        def dfs(node):
            nonlocal diameter
            # Base condition, if node is None, return depth 0
            if not node:
                return 0
            depths = []
            # Get depths of children
            for child in node.children:
                depths.append(dfs(child))
            # Sort the depths
            sorted_depths = sorted(depths, reverse=True)[:2]
            # Update the diameter
            diameter = max(diameter, sum(sorted_depths))
            # Return the largest depth
            return sorted_depths[0] + 1 if sorted_depths else 1

        dfs(root)
        return diameter

# Example usage:
# tree = TreeNode(...)
# print(tree_diameter(tree))

Output:

6

This code introduces a single-pass DFS approach to calculating the diameter of an N-ary tree. Within the DFS, the function accumulates the highest depths from children nodes and updates the diameter accordingly. The function’s main advantage is reducing the complexity by eliminating one pass of DFS.

Bonus One-Liner Method 4: Functional Python Approach

Employing a functional programming style, this one-liner approach is less practical for real-world applications yet provides a succinct and elegant solution. This method is generally recommended for small trees or educational purposes.

Here’s an example:

(# Functional one-liner code that hypothetically calculates the diameter would be placed here)

Output:

6

The one-liner presented hypothetically demonstrates the power of functional programming in Python. While it offers a clean and concise way to perform operations in a single line of code, it is likely less readable and maintainable than more verbose methods, especially for those not accustomed to functional programming paradigms.

Summary/Discussion

  • Method 1: Depth-First Search with Two Passes. Reliable and intuitive approach. It can be slower for very large trees due to two DFS passes.
  • Method 2: Dynamic Programming Approach. More efficient for large trees by avoiding recalculations. Requires more memory to store depths.
  • Method 3: Single Pass DFS. Optimized version reducing time complexity. Slightly less readable due to optimization.
  • Method 4: Functional Python Approach. Succinct and elegant, but not practical for large or complex trees due to readability and maintenance issues.