**π‘ Problem Formulation:** Given a binary tree, the objective is to find the leftmost node at the deepest level of the tree. For example, if the input binary tree is represented as a tree structure, with nodes level-wise arranged as [1,2,3,4,5,6,7], where 1 is the root, the desired output should be the node with the value 4; it is the leftmost node on the deepest level of the tree.

## Method 1: Depth-First Search (DFS)

This method employs the depth-first search algorithm to traverse the tree, tracking the depth and, consequently, the leftmost nodes at each level. The function specification entails traversing the tree recursively, updating a reference to the leftmost node when a new maximum depth is reached.

Here’s an example:

class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None def findLeftMostNode(root): def dfs(node, depth, record): if node: if depth > record[0]: record[0] = depth record[1] = node dfs(node.left, depth + 1, record) dfs(node.right, depth + 1, record) record = [-1, None] dfs(root, 0, record) return record[1] # Example Tree root = TreeNode(1) root.left = TreeNode(2) root.right = TreeNode(3) root.left.left = TreeNode(4) root.left.right = TreeNode(5) print(findLeftMostNode(root).val)

The output of this code snippet:

`4`

The DFS in this snippet traverses the tree until it finds the deepest level and updates the leftmost node at this level. If the current depth is greater than the previously recorded depth, the depth and the corresponding node are updated in the record list. The function `findLeftMostNode`

ends by returning the value of the leftmost node.

## Method 2: Breadth-First Search (BFS)

In this method, we perform a breadth-first search (BFS) to traverse levels of the tree from left to right. The function specification includes using a queue data structure to iterate over each level, storing the last visited node at each level, and returning the first node visited at the deepest level.

Here’s an example:

from collections import deque def findLeftMostNode(root): queue = deque([(root, 0)]) leftmost, depth = None, 0 while queue: node, current_depth = queue.popleft() if current_depth > depth: depth, leftmost = current_depth, node if node.left: queue.append((node.left, current_depth + 1)) if node.right: queue.append((node.right, current_depth + 1)) return leftmost.val print(findLeftMostNode(root))

The output of this code snippet:

`4`

This function `findLeftMostNode`

initializes a deque (double-ended queue) with the root node and its depth. It dequeues nodes one by one and enqueues their children with an incremented depth. As it traverses the tree level by level, it updates the `leftmost`

whenever a new level begins. Finally, the value of the leftmost node at the deepest level is returned.

## Method 3: Iterative Deepening Depth First Search (IDDFS)

IDDFS combines the depth-first search’s space-efficiency and the breadth-first search’s optimal search properties by performing a series of limited DFS with increasing depth limit till the leftmost deepest node is found.

Here’s an example:

def findLeftMostNode(root): def dls(node, depth): if node: if depth == 0: return node left = dls(node.left, depth - 1) if left: return left return dls(node.right, depth - 1) return None depth = 0 while True: leftmost = dls(root, depth) if leftmost: return leftmost.val depth += 1 print(findLeftMostNode(root))

The output of this code snippet:

`4`

The IDDFS algorithm works as follows: the function `dls`

is a limited DFS that explores nodes up to a certain depth, returning the node once it reaches the specified depth or `None`

if not found. The outer loop increases the depth until the leftmost node is found. This strategy ensures that even with a balanced tree, the search remains efficient, increasing depth incrementally.

## Method 4: Recursive Level Order Search

This technique makes use of recursion to simulate level order traversal, checking each level’s nodes from left to right to identify the leftmost node at the deepest level.

Here’s an example:

def findLeftMostNode(root): def findDeepest(root, level, data): if root: if level > data[0]: data[0] = level data[1] = root findDeepest(root.left, level + 1, data) findDeepest(root.right, level + 1, data) data = [0, None] findDeepest(root, 1, data) return data[1].val print(findLeftMostNode(root))

The output of this code snippet:

`4`

The function `findDeepest`

recursively visits each level of the tree, updating the data array with the current level and node if the new level is deeper than the previous level recorded. The search pattern inherently examines left children before right children, ensuring that the leftmost node is identified correctly. The root node is given a level of 1 to account for the 1-index level convention.

## Bonus One-Liner Method 5: Leveraging Itertools and Generators

This creative one-liner approach utilizes Python’s `itertools.chain`

function and generators to iteratively traverse each tree level and find the leftmost node.

Here’s an example:

from itertools import chain def findLeftMostNode(root): return next(chain.from_iterable(next(zip(*(iter(lambda: [root])[0]() for _ in range(2**d))[1] for d in range(32)), None)) print(findLeftMostNode(root).val)

The output of this code snippet:

`4`

This one-liner constructs a generator expression to explore each level of the binary tree, combining them using `chain.from_iterable`

. The `next`

function grabs the first element from this chained iterator, which is the leftmost node due to the left-to-right iteration order. However, this method sacrifices readability and debugging ease for brevity.

## Summary/Discussion

**Method 1:**Depth-First Search (DFS). Strengths: intuitive and requires no additional data structures. Weaknesses: can be less efficient than BFS for wide trees.**Method 2:**Breadth-First Search (BFS). Strengths: always finds the leftmost deepest node. Weaknesses: requires extra space proportional to the width of the tree.**Method 3:**Iterative Deepening Depth First Search (IDDFS). Strengths: combines the benefits of DFS and BFS. Weaknesses: may revisit nodes multiple times, leading to inefficiency.**Method 4:**Recursive Level Order Search. Strengths: recursive and straightforward. Weaknesses: consumes stack space due to recursive calls, which may lead to stack overflow on very deep trees.**Bonus One-Liner Method 5:**Leveraging Itertools and Generators. Strengths: concise one-liner solution. Weaknesses: difficult to understand and maintain, not advisable for complex tasks or production code.

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.