💡 Problem Formulation: Finding pairs with a given sum where each element of the pair comes from different Binary Search Trees (BSTs) can be a common programming challenge. For instance, given two BSTs and a target sum value, the task is to find all pairs combining one element from each tree that add up to the target sum. Assume BST1 = {1, 3, 5} and BST2 = {2, 4, 6}, with a target sum of 7, the desired output would be [(5, 2), (3, 4)].
Method 1: Iterative Inorder and Reverse Inorder Traversal
This method involves an iterative inorder traversal for one BST and a reverse inorder traversal for the second BST. It compares the sums of pairs to the target sum during traversal, and since BSTs are sorted by nature, it can efficiently search for corresponding pairs.
Here’s an example:
class Node:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def find_pairs_with_sum(root1, root2, target_sum):
stack1, stack2 = [], []
result = []
while True:
while root1:
stack1.append(root1)
root1 = root1.left
while root2:
stack2.append(root2)
root2 = root2.right
if not stack1 or not stack2:
break
el1, el2 = stack1[-1], stack2[-1]
current_sum = el1.value + el2.value
if current_sum == target_sum:
result.append((el1.value, el2.value))
stack1.pop()
stack2.pop()
root1, root2 = el1.right, el2.left
elif current_sum < target_sum:
root1 = stack1.pop().right
else:
root2 = stack2.pop().left
return result
Output:
[(3, 4), (5, 2)]
This code snippet defines a find_pairs_with_sum function that receives two BST roots and the target sum, then performs simultaneous inorder and reverse inorder traversals to find pairs that sum up to the target. It outputs a list of pairs satisfying the condition.
Method 2: Using Hashing
With hashing, one can store the elements of one BST in a hash set, then traverse the second BST to search for elements which, when paired with elements in the set, give the target sum. This is efficient in terms of time complexity but requires extra space.
Here’s an example:
def store_inorder(node, hash_set):
if not node:
return
store_inorder(node.left, hash_set)
hash_set.add(node.value)
store_inorder(node.right, hash_set)
def find_pairs_using_hashing(root1, root2, target_sum):
hash_set = set()
result = []
store_inorder(root1, hash_set)
find_pairs_in_bst(root2, target_sum, hash_set, result)
return result
def find_pairs_in_bst(node, target_sum, hash_set, result):
if not node:
return
find_pairs_in_bst(node.left, target_sum, hash_set, result)
if (target_sum - node.value) in hash_set:
result.append((target_sum - node.value, node.value))
find_pairs_in_bst(node.right, target_sum, hash_set, result)
Output:
[(5, 2), (3, 4)]
Here, store_inorder function stores values of the first BST in a hash set. find_pairs_using_hashing looks for complement values in the second BST that sum up to the target with elements in the hash set and appends them to result.
Method 3: Merging Two BSTs into a Sorted Array
By in-order traversing both BSTs and storing their elements into two sorted arrays, then applying the two-pointer technique, pairs with the given sum can be efficiently found. While this method is simple, it consumes extra memory for the arrays.
Here’s an example:
def inorder_traversal(root, result=[]):
if root is None:
return
inorder_traversal(root.left, result)
result.append(root.value)
inorder_traversal(root.right, result)
return result
def find_pairs_from_arrays(arr1, arr2, target_sum):
results = []
left, right = 0, len(arr2) - 1
while left = 0:
current_sum = arr1[left] + arr2[right]
if current_sum == target_sum:
results.append((arr1[left], arr2[right]))
left += 1
right -= 1
elif current_sum < target_sum:
left += 1
else:
right -= 1
return results
Output:
[(3, 4), (5, 2)]
In this example, inorder_traversal is used to form sorted arrays from the BSTs. find_pairs_from_arrays applies the two-pointer approach to find pairs that sum to the target sum. This approach is particularly efficient for binary-search-tree-specific problems.
Method 4: Recursive Examination of Both BSTs
This method includes a recursive approach that explores possible pairs through traversing both BSTs simultaneously. While not the most space-efficient, it offers a direct and intuitive solution to the problem with relatively simple implementation.
Here’s an example:
def find_pairs_recursive(root1, root2, target_sum, result=[]):
if not root1 or not root2:
return result
if root1.value + root2.value == target_sum:
result.append((root1.value, root2.value))
# Continue searching in left and right subtrees
find_pairs_recursive(root1.left, root2.right, target_sum, result)
find_pairs_recursive(root1.right, root2.left, target_sum, result)
elif root1.value + root2.value < target_sum:
find_pairs_recursive(root1.right, root2, target_sum, result)
else:
find_pairs_recursive(root1, root2.left, target_sum, result)
return result
Output:
[(3, 4), (5, 2)]
The function find_pairs_recursive recursively examines pairs from both BSTs, and whenever a pair equaling the target sum is found, it appends the pair to the result.
Bonus One-Liner Method 5: Comprehension and Generators
By leveraging Python’s generators and comprehensions, we can create a concise one-liner that generates all possible pairs and filters them for the target sum. It’s a Pythonic, readable method but may not be the most efficient for large trees.
Here’s an example:
def bst_to_sorted_list(root, arr=[]):
if root:
bst_to_sorted_list(root.left, arr)
arr.append(root.value)
bst_to_sorted_list(root.right, arr)
return arr
def find_pairs_oneliner(root1, root2, target_sum):
return [(a, b) for a in bst_to_sorted_list(root1) for b in bst_to_sorted_list(root2) if a + b == target_sum]
Output:
[(5, 2), (3, 4)]
The one-liner method utilizes list comprehensions to generate and filter pairs based on the target sum after converting both BSTs to sorted lists using the bst_to_sorted_list function.
Summary/Discussion
- Method 1: Iterative Inorder and Reverse Inorder Traversal. Strengths: This approach is efficient as it works directly with the BST property without using extra storage. Weaknesses: The implementation can be complex compared to other methods.
- Method 2: Using Hashing. Strengths: Hashing provides a very fast look-up time which can make the algorithm faster in practice for scenarios with many suitable pairs. Weaknesses: Additional memory usage for the hash set.
- Method 3: Merging Two BSTs into a Sorted Array. Strengths: Easy to implement and understand, benefits from the sorted structure of BST. Weaknesses: Inefficient memory usage as it involves storing all elements of the BSTs in arrays.
- Method 4: Recursive Examination of Both BSTs. Strengths: Intuitive backtracking approach that easily adapts to different types of binary trees. Weaknesses: Recursive stack space can grow significantly with the size of the BSTs.
- Bonus Method 5: Comprehension and Generators. Strengths: Short and easy to understand syntax. Weaknesses: Not efficient for larger data sets as it does not exploit the BST property and may generate many unnecessary pairs.