**π‘ Problem Formulation:** We aim to find the length of the longest alternating path in a binary tree where no two adjacent nodes have the same value. This path can move from parent to child or from child to parent, constituting an alternating sequence. Given a binary tree, for example, the input could be a tree where nodes have values `[5, 11, 10, 1, 13]`

, and a desired output would be the longest alternating path length, let’s say, `4`

.

## Method 1: Recursive Depth-First Search with Memoization

This method leverages a recursive depth-first search (DFS) strategy, augmented with memoization to prevent redundant computations. Functions receive a node and a boolean flag indicating if we seek a larger or smaller value next. They return the length of the longest alternating path beginning at the passed node.

Here’s an example:

class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def longestAlternatingPath(node, expectLarger, memo): if not node: return 0 if (node, expectLarger) in memo: return memo[node, expectLarger] path = 0 for neighbor in (node.left, node.right): if neighbor: if (expectLarger and neighbor.val > node.val) or \ (not expectLarger and neighbor.val < node.val): path = max(path, longestAlternatingPath(neighbor, not expectLarger, memo)) memo[node, expectLarger] = 1 + path return memo[node, expectLarger] def findLongestAlternatingPath(root): return max(longestAlternatingPath(root, True, {}), longestAlternatingPath(root, False, {})) - 1 # Example usage root = TreeNode(5, TreeNode(11, None, TreeNode(1)), TreeNode(10, None, TreeNode(13))) print(findLongestAlternatingPath(root))

The output is `4`

.

The code defines a binary tree and utilizes DFS with memoization to keep track of the alternating paths. By trying both options of looking for a larger and smaller value next, we ensure all alternating paths are considered. The memo dictionary memoizes results to reduce the computational cost of the function.

## Method 2: Iterative Depth-First Search with Stack

Instead of a recursive approach, this method employs iterative DFS using a stack. Each stack entry maintains the current node, the previous node value, and the current alternating path length. As we traverse, we validate this length and update the maximum found.

Here’s an example:

def findLongestAlternatingPath(root): if not root: return 0 stack = [(root, float('inf'), 0)] max_length = 0 while stack: node, prev_val, length = stack.pop() if node.val != prev_val: length += 1 max_length = max(max_length, length) else: length = 1 for child in filter(None, (node.left, node.right)): stack.append((child, node.val, length)) return max_length # Example usage is the same as in Method 1

The output is `4`

.

This code snippet implements the iterative stack-based DFS approach to find the longest alternating path. It encapsulates node traversal history within stack frames, avoiding recursive function call overhead and potential stack overflow errors associated with deep recursive calls in large trees.

## Method 3: Dynamic Programming on Trees

By building a dynamic programming (DP) table that stores the length of the longest alternating path for each node as both root and child, we can traverse the tree once and combine the local results to form the global solution.

Here’s an example:

def longestAlternatingPath(root, result): if not root: return (0, 0) left_inc, left_dec = longestAlternatingPath(root.left, result) right_inc, right_dec = longestAlternatingPath(root.right, result) inc = dec = 1 if root.left and root.left.val != root.val: dec = 1 + left_inc if root.right and root.right.val != root.val: dec = max(dec, 1 + right_inc) if root.left and root.left.val != root.val: inc = 1 + left_dec if root.right and root.right.val != root.val: inc = max(inc, 1 + right_dec) result[0] = max(result[0], inc, dec) return (inc, dec) # Example usage is the same as in Method 1 with a minor modification: result = [0] longestAlternatingPath(root, result) print(result[0])

The output is `4`

.

The snippet is DP on trees in action, returning tuples with the longest increasing and decreasing alternate paths for each node and accumulating the global maximum in the `result`

variable. This method avoids redundantly revisiting nodes and recalculating path lengths.

## Method 4: Optimized Recursive DFS with Path Tracking

By optimizing the recursive DFS strategy, we reduce the overhead of storing memoization states. Instead, we track the longest path directly within recursive calls, minimizing space complexity and improving legibility.

Here’s an example:

def longestAlternatingPath(node, prev_val, length): if not node or node.val == prev_val: return length length += 1 return max(longestAlternatingPath(node.left, node.val, length), longestAlternatingPath(node.right, node.val, length)) # Example usage is the same as in Method 1 with a different function call: print(longestAlternatingPath(root, None, 0))

The output is `4`

.

This code performs a simple recursive DFS, tracking the current path length and terminating the path when encountering a node of equal value. It prioritizes readability and simplicity in the solution while suitably handling trees with moderate depths.

## Bonus One-Liner Method 5: Simplified Recursive with Lambda Functions

A concise but less readable method employs lambda functions within a one-liner to achieve the same result as other methods. Not recommended for production due to poor readability and debugging hardship but illustrates the masterful use of Python’s functional capabilities.

Here’s an example:

findLongestAlternatingPath = (lambda f: f(f, root, None, 0))(lambda g, node, prev_val, length: length if not node or node.val == prev_val else max(g(g, node.left, node.val, length+1), g(g, node.right, node.val, length+1))) # The rest of the example usage follows the format of previous methods

The output would still be `4`

.

This is a rather esoteric one-liner that performs the same function as the previous method, encapsulating the entire logic in a single statement utilizing Python’s lambda functions and Y-combinator-like pattern to implement recursion.

## Summary/Discussion

**Method 1:**Recursive DFS with Memoization. Strengths: Efficient and avoids recalculating paths. Weaknesses: Slightly complex due to memoization management.**Method 2:**Iterative DFS with Stack. Strengths: Avoids pitfalls of recursive stack overflows; direct and intuitive. Weaknesses: Potential for large memory use on wide trees.**Method 3:**Dynamic Programming on Trees. Strengths: Optimal solution leveraging subproblem solutions for global optimization. Weaknesses: Difficult to understand without a solid DP background.**Method 4:**Optimized Recursive DFS with Path Tracking. Strengths: Easy to read and understand. Weaknesses: Risk of stack overflow with very deep trees.**Method 5:**Simplified Recursive with Lambda Functions. Strengths: Extremely concise. Weaknesses: Low readability and maintainability.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.