**π‘ Problem Formulation:** We need a way to verify in a binary tree whether each non-leaf node’s value is indeed the sum of its children’s values. For instance, given a binary tree where a non-leaf node with a value of 10 has two children with values 4 and 6, our check should return true as 10 equals 4 plus 6.

## Method 1: Recursive Approach

This method involves a recursive function that traverses the tree. If the node is null or a leaf, it returns true. Otherwise, it checks if the node’s value is the sum of its childrenβs values and then recursively checks the children.

Here’s an example:

class Node: def __init__(self, value): self.value = value self.left = None self.right = None def check_sum(node): if node is None or (node.left is None and node.right is None): # Leaf or null check return True left_val = node.left.value if node.left else 0 right_val = node.right.value if node.right else 0 return (node.value == left_val + right_val) and check_sum(node.left) and check_sum(node.right) # Example Tree root = Node(10) root.left = Node(4) root.right = Node(6) print(check_sum(root))

The output of this code is: `True`

.

This snippet defines a simple binary tree and a function `check_sum`

to recursively verify that each node value, except the leaves, is the sum of its children’s values. It’s a clear and simple recursive descent that checks the required property.

## Method 2: Iterative Depth-First Search (DFS)

Iterative DFS uses a stack to simulate the call stack of a recursive approach. It’ll iterate over each node and check the sum condition until the stack is empty.

Here’s an example:

class Node: def __init__(self, value): self.value = value self.left = None self.right = None def check_sum_iterative(root): if not root: return True stack = [root] while stack: node = stack.pop() if node.left is None and node.right is None: continue child_sum = 0 if node.left: child_sum += node.left.value stack.append(node.left) if node.right: child_sum += node.right.value stack.append(node.right) if node.value != child_sum: return False return True # Example Tree root = Node(10) root.left = Node(4) root.right = Node(6) print(check_sum_iterative(root))

The output of this code is: `True`

.

Instead of recursion, this method uses a stack to keep track of the nodes to visit. It iteratively performs the same check as the recursive method, proving its versatility across different implementations.

## Method 3: Breadth-First Search (BFS)

Breadth-First Search algorithm uses a queue to traverse the tree by levels from top to bottom. It checks each node’s value against the sum of its children during traversal.

Here’s an example:

from collections import deque class Node: def __init__(self, value): self.value = value self.left = None self.right = None def check_sum_bfs(root): if not root: return True queue = deque([root]) while queue: node = queue.popleft() if node.left is None and node.right is None: continue child_sum = 0 if node.left: child_sum += node.left.value queue.append(node.left) if node.right: child_sum += node.right.value queue.append(node.right) if node.value != child_sum: return False return True # Example Tree root = Node(10) root.left = Node(4) root.right = Node(6) print(check_sum_bfs(root))

The output of this code is: `True`

.

This code snippet performs a level order traversal using a queue to check if each node follows our sum condition. It ensures all nodes satisfy the constraint before returning `True`

.

## Method 4: Utilizing a Helper Function

This method uses a helper function to abstract some of the logic within the recursive approach, potentially improving readability and making it easier to manage complex logic or additional functionality.

Here’s an example:

class Node: def __init__(self, value): self.value = value self.left = None self.right = None def is_sum_node(node): left_val = node.left.value if node.left else 0 right_val = node.right.value if node.right else 0 return node.value == left_val + right_val def check_sum_with_helper(node): if node is None or (node.left is None and node.right is None): return True if not is_sum_node(node): return False return check_sum_with_helper(node.left) and check_sum_with_helper(node.right) # Example Tree root = Node(10) root.left = Node(4) root.right = Node(6) print(check_sum_with_helper(root))

The output of this code is: `True`

.

In this version, the `is_sum_node`

helper function simplifies the process of checking a nodeβs value against its childrenβs sum. The `check_sum_with_helper`

function maintains the structure of the recursive method.

## Bonus One-Liner Method 5: Using Lambda Expressions

For the enthusiasts of concise code, Python lambdas can sometimes pack the logic into a one-liner function. This is less readable but can be fun for code golfing.

Here’s an example:

class Node: def __init__(self, value): self.value = value self.left = None self.right = None check_sum_lambda = lambda n: n is None or ((n.left is None and n.right is None) or (n.value == (n.left.value if n.left else 0) + (n.right.value if n.right else 0) and check_sum_lambda(n.left) and check_sum_lambda(n.right))) # Example Tree root = Node(10) root.left = Node(4) root.right = Node(6) print(check_sum_lambda(root))

The output of this code is: `True`

.

This snippet shows a one-liner lambda function that acts as a drop-in replacement for our recursive sum check method. While compact, it is recommended for smaller and simpler checks due to legibility concerns.

## Summary/Discussion

**Method 1:**Recursive Approach. Straightforward and elegant. It might hit the recursion depth limit for very deep trees.**Method 2:**Iterative DFS. Provides a non-recursive solution that avoids potential stack overflow issues with recursion, but can be less intuitive.**Method 3:**BFS. It traverses the tree level-by-level and may be more intuitive for some. Requires additional space for the queue proportional to the breadth of the tree.**Method 4:**Helper Function. Encapsulates node sum check logic, which can make the code cleaner and more modular. Adds a bit of overhead but improves readability.**Method 5:**Lambda Expressions. A compact solution that’s not as easily readable. Useful for those preferring concise code but may be harder to maintain.

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.