**π‘ Problem Formulation:** Given two binary trees, we aim to determine if they are identical in both structure and content. Identical trees must have the same number of nodes with identical positions and corresponding nodes must contain the same value. For instance, if Tree A and Tree B both have a root node with the value ‘3’ and left child node ‘4’, they are identical only if the entire structure and node values of both trees match exactly.

## Method 1: Recursive Traversal

The recursive traversal method involves comparing nodes recursively, checking each node’s value simultaneously in both trees, and ensuring both left and right subtrees are also identical. This function should return `true`

if both trees are the same or `false`

otherwise. The function is called recursively for left and right subtrees.

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 is_same_tree(p, q): if not p and not q: return True if not p or not q or p.val != q.val: return False return is_same_tree(p.left, q.left) and is_same_tree(p.right, q.right) # Example Trees tree1 = TreeNode(1, TreeNode(2), TreeNode(3)) tree2 = TreeNode(1, TreeNode(2), TreeNode(3)) print(is_same_tree(tree1, tree2))

Output:

True

This code snippet defines a binary tree using the TreeNode class and checks if two binary trees (tree1 and tree2) are identical using the `is_same_tree()`

function. It performs a depth-first traversal, comparing nodes from both trees and ensuring that the structures and values are the same at each step.

## Method 2: Iterative Traversal

Iterative traversal is similar to recursive traversal, but uses a stack to compare nodes without the function call overhead. We traverse both trees in parallel and compare nodes as they’re visited. This can be done using any traversal order, but here we’ll demonstrate pre-order traversal.

Here’s an example:

def is_same_tree_iterative(p, q): stack = [(p, q)] while stack: node1, node2 = stack.pop() if not node1 and not node2: continue if not node1 or not node2 or node1.val != node2.val: return False stack.append((node1.right, node2.right)) stack.append((node1.left, node2.left)) return True # Assume TreeNode class and example trees as defined in Method 1 print(is_same_tree_iterative(tree1, tree2))

Output:

True

The code snippet here implements an iterative version of the `is_same_tree()`

function called `is_same_tree_iterative()`

. It uses a stack to iteratively traverse both trees and perform node comparisons, maintaining structure and value integrity checks.

## Method 3: Level-Order Traversal

This approach uses level-order traversal to compare nodes’ values at each level of both trees using a queue. It is a breadth-first search strategy that visits each level of the tree before proceeding to the next one. This method is intuitive and closely resembles how one might visually compare two trees.

Here’s an example:

from collections import deque def is_same_tree_level_order(p, q): queue = deque([(p, q)]) while queue: for _ in range(len(queue)): node1, node2 = queue.popleft() if not node1 and not node2: continue if not node1 or not node2 or node1.val != node2.val: return False queue.append((node1.left, node2.left)) queue.append((node1.right, node2.right)) return True # Assume TreeNode class and example trees as defined in Method 1 print(is_same_tree_level_order(tree1, tree2))

Output:

True

The code snippet illustrates `is_same_tree_level_order()`

, a function that uses level-order traversal to compare trees. Nodes are visited level by level, and their existence and values are compared using a queue, with the process continuing as long as the queue isn’t empty.

## Method 4: String Serialization

In this method, we serialize both trees to a string representation and then compare these strings. Serialization involves a systematic encoding of tree nodes and null markers to fully represent the tree’s structure and values. We can use pre-order traversal for serialization.

Here’s an example:

def serialize(root): return "X" if not root else f"{root.val} {serialize(root.left)} {serialize(root.right)}" # Assume TreeNode class and example trees as defined in Method 1 print(serialize(tree1) == serialize(tree2))

Output:

True

The code snippet creates a serialization of each tree using the `serialize()`

function, which transforms the tree into a string using pre-order traversal. By comparing the resulting strings for both trees, we can determine if they are identical.

## Bonus One-Liner Method 5: Using Equality Operator with Nested Tuples

A Pythonic one-liner can achieve the comparison task by converting the tree structure into nested tuples and then using the equality operator to compare them. This method leverages the tuple data structure’s innate ability to be compared for equality, which includes recursive checks for nested structures.

Here’s an example:

def tupleize(root): return (root.val, tupleize(root.left), tupleize(root.right)) if root else None # Assume TreeNode class and example trees as defined in Method 1 print(tupleize(tree1) == tupleize(tree2))

Output:

True

This code snippet uses the `tupleize()`

function to convert each tree into a nested tuple reflecting its structure and value. The comparison is then a simple check using the ‘==’ operator.

## Summary/Discussion

**Method 1: Recursive Traversal.**Uses simple recursion for depth-first search. Strengths: easy to understand and implement. Weaknesses: possible stack overflow for deep trees.**Method 2: Iterative Traversal.**Uses explicit stack to avoid recursion overhead. Strengths: better control over traversal, no recursion stack limit. Weaknesses: slightly more complex implementation.**Method 3: Level-Order Traversal.**Uses a queue for breadth-first search. Strengths: intuitive comparison level by level. Weaknesses: requires additional memory proportional to the width of the tree.**Method 4: String Serialization.**Compares serialized string representations of trees. Strengths: simple string comparison, good for debugging. Weaknesses: additional overhead of serialization and more memory usage.**Bonus Method 5: Using Equality Operator with Nested Tuples.**Pythonic, concise one-liner with implicit recursion. Strengths: elegant and efficient. Weaknesses: less readable, requires understanding of tuple comparison.

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.