Two Sum IV - Input is a BST

Given the root of a binary search tree (BST) and an integer ( k ), the task is to determine if there exist two elements in the BST such that their sum is equal to ( k ).

Algorithm

  1. Convert the BST to a List: Implement a function that performs an in-order traversal of the BST and appends the value of each node to a list. The in-order traversal will produce a sorted list since it’s a BST.

  2. Use Two Pointers Technique: Initialize two pointers, one at the start and the other at the end of the sorted list. Then, iterate until the pointers meet:

    a. Calculate the Sum: Calculate the sum of the elements at the positions of the two pointers.

    b. Compare with Target: If the sum is equal to ( k ), return True. If the sum is less than ( k ), move the left pointer to the right. If the sum is greater than ( k ), move the right pointer to the left.

  3. Return False: If the iteration completes without finding a pair that sums to ( k ), return False.

Code

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class Solution:
    def findTarget(self, root: Optional[TreeNode], k: int) -> bool:
        def inOrderTraversal(node):
            if not node:
                return []
            return inOrderTraversal(node.left) + [node.val] + inOrderTraversal(node.right)

        sorted_list = inOrderTraversal(root)
        left, right = 0, len(sorted_list) - 1

        while left < right:
            current_sum = sorted_list[left] + sorted_list[right]
            if current_sum == k:
                return True
            elif current_sum < k:
                left += 1
            else:
                right -= 1

        return False

Explanation

  • The recursive function inOrderTraversal is used to perform an in-order traversal of the BST, converting it into a sorted list.
  • The two pointers technique is applied to the sorted list, searching for a pair of elements that sum to ( k ).

Insights

  • Time Complexity: ( O(n) ), where ( n ) is the number of nodes in the BST. We traverse the tree once to create the sorted list, and the two pointers technique takes ( O(n) ) time.
  • Space Complexity: ( O(n) ), to store the sorted list.

This approach efficiently checks if there exist two elements in the given BST whose sum is equal to the target value ( k ), leveraging the properties of a BST and the two pointers technique.

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# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} root
# @param {Integer} k
# @return {Boolean}
def two_sum(array, target)
  i = 0
  j = array.size - 1
  
  while i < j
    sum = array[i] + array [j]

    if sum == target
      return true
    elsif sum < target
      i += 1
    else
      j -= 1
    end
  end
  
  return false
end

def inorder(node, values)
  if node.nil?
    return
  end
  
  inorder(node.left, values)
  values << node.val
  inorder(node.right, values)
end

def find_target(root, k)  
  values = []
  inorder(root, values)
  two_sum(values, k)    
end

Identifying Problem Isomorphism

An approximate isomorphism: “Find two non-overlapping sub-arrays each with target sum”.

In “Two Sum IV - Input is a BST”, you are given a binary search tree and a target number, and your task is to find two nodes in the tree such that their values add up to the target number.

“Find two non-overlapping sub-arrays each with target sum” shares similarities. In this problem, you are given an array and a target sum. The goal is to find two non-overlapping sub-arrays such that the sum of the elements in each sub-array is equal to the target.

In both problems, you need to find two distinct groups (whether they are nodes in a tree or sub-arrays) that each contribute to a target sum. The methods to solve them are also similar. You can use a HashMap to track the potential values that can form the target sum in both cases.

“Two Sum IV - Input is a BST” is simpler due to its structure. Binary Search Trees have the property that the left child is smaller and the right child is larger than the parent node, which allows us to use a two-pointer technique to solve the problem. On the other hand, “Find two non-overlapping sub-arrays each with target sum” is a bit more complex as the array doesn’t have the same property as BST, so a two-pointer approach won’t work directly here. We would need to use a dynamic programming approach.

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class Solution:
    def findTarget(self, root: TreeNode, k: int) -> bool:
        vals = {}

        def helper(node):
            if not node:
                return False

            if node.val in vals:
                return True

            vals[k - node.val] = True

            return helper(node.left) or helper(node.right)

        return helper(root)

Problem Classification

This problem falls into the category Search Algorithms. The primary reason for this categorization is the involvement of a Binary Search Tree (BST), which is a specific type of data structure, and the need to search for two elements within this data structure that satisfy a certain condition.

‘What’ Components:

  1. Binary Search Tree (BST): A binary search tree is a specific type of binary tree where each node has a unique value. For each node, the nodes to its left are smaller, and the nodes to its right are larger.

  2. Integer ‘k’: This is a target sum that we are looking to match by adding two elements from the BST.

  3. Return True or False: The task is to return a boolean value - ‘True’ if there exist two elements in the BST whose sum is equal to ‘k’, and ‘False’ otherwise.

This problem can be classified as a Search Problem, specifically a Two-Pointer Search Problem in a BST. It involves finding two elements in the BST such that their sum equals a given target ‘k’. The structure of the BST allows for an efficient search strategy, namely in-order traversal, which returns elements in ascending order and enables a two-pointer search technique. The task involves a search operation that has to be conducted following the properties of the BST, thus making this classification appropriate.

Language Agnostic Coding Drills

  1. Dissecting the Code

Here are the distinct coding concepts found in the code:

  • Class Definition and Method Creation: The code is structured in a class and uses a method to solve the problem.

  • Dictionary Usage: A dictionary is used to keep track of the values needed to reach the target sum from each node.

  • Binary Tree Traversal: The solution involves a recursive depth-first traversal of the binary tree.

  • Boolean Evaluation: The solution requires evaluating whether certain conditions are true or false.

  • Recursion: The function calls itself to traverse the binary tree.

  1. Coding Concepts Listed in Order of Increasing Difficulty

    1. Class Definition and Method Creation: This is a basic concept that involves structuring code in an organized manner. It is foundational to most object-oriented programming languages.

    2. Dictionary Usage: While still fundamental, understanding dictionaries (or equivalent structures like maps or hash tables in other languages) requires a bit more understanding of data structures and how to manipulate them.

    3. Boolean Evaluation: This involves understanding conditions and logic, which is a bit more complex than simply understanding data structures.

    4. Recursion: Recursion is a more advanced concept that involves a function calling itself. It requires a solid understanding of the call stack and how functions work.

    5. Binary Tree Traversal: This concept is specific to binary trees, a more advanced data structure, and requires understanding of recursion and the properties of binary trees.

  2. Problem-Solving Approach

The initial step is to set up the class and method. We also initialize a dictionary to store the values we need to reach our target sum.

We then begin a depth-first traversal of the binary tree. For each node we visit, we check whether its value is already in our dictionary. If it is, that means we’ve found two numbers in our tree that sum to our target ‘k’, so we return ‘True’.

If the node’s value isn’t in the dictionary, we add the difference between our target and the node’s value to the dictionary. This is the value we’re looking for as we continue our traversal.

We repeat this process for every node in the tree, performing a depth-first search by visiting each node’s left child and then its right child.

Finally, we return ‘False’ if we’ve traversed the entire tree without finding two numbers that sum to ‘k’.

Targeted Drills in Python

  1. Python-based Coding Drills for Each Identified Concept

a. Class Definition and Method Creation

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class MyClass:
    def my_method(self, param):
        print(f"Method called with parameter: {param}")
        
obj = MyClass()
obj.my_method("Hello World")

b. Dictionary Usage

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my_dict = {"apple": 1, "banana": 2}
print(my_dict["apple"]) # Output: 1
my_dict["cherry"] = 3
print(my_dict) # Output: {"apple": 1, "banana": 2, "cherry": 3}

c. Boolean Evaluation

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num = 10
is_even = num % 2 == 0
print(is_even) # Output: True

d. Recursion

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def recursive_sum(n):
    if n == 0:
        return 0
    else:
        return n + recursive_sum(n-1)

print(recursive_sum(5)) # Output: 15

e. Binary Tree Traversal (DFS)

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class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

def dfs(node):
    if node is None:
        return
    print(node.val)
    dfs(node.left)
    dfs(node.right)

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)

dfs(root) # Output: 1 2 3
  1. Problem-Specific Concepts
  • Tracking values in dictionary and checking for their existence

For this problem, the essential concept is understanding how to use a dictionary to track the values we have seen and the values we are looking for. We add the difference between the target sum and the current node’s value to the dictionary. This is how we track what we’re looking for as we continue to traverse the tree. The presence of a node’s value in the dictionary signifies the existence of two numbers in the tree that add up to the target.

  1. Integration of Drills to Form Final Solution

The first step in solving this problem involves defining a class and creating a method within it. This setup provides a blueprint for the solution.

The next step involves setting up a dictionary which we will use to track the values we have seen and the values we are looking for.

We then initiate a binary tree traversal using recursion. This is where we integrate the other drills. For each node, we perform a boolean evaluation to check if its value is in the dictionary. If it is, we return True. If not, we add to the dictionary the difference between our target and the node’s value.

The recursion continues until we have traversed the entire tree. If we do not find any two numbers that sum up to the target, we return False.

The core concept here is how we use the dictionary to keep track of the values we need to reach our target sum, along with the binary tree traversal to visit each node. It’s the combination of these drills that allows us to solve the problem.