Problem Description

You are given a binary search tree (BST), a target value, and an integer k. Your task is to find the k values in the BST that are closest to the target value.

The problem requires you to:

• Traverse the BST and examine all node values
• Calculate the distance (absolute difference) between each node value and the target
• Return exactly k values that have the smallest distances to the target
• The values can be returned in any order

Key points about the problem:

• The tree is a valid BST, meaning left subtree values are less than the root, and right subtree values are greater than the root
• The target can be any floating-point number (not necessarily in the tree)
• You are guaranteed that there is exactly one unique set of k closest values (no ties at the k-th position)

For example, if you have a BST with values [1, 2, 3, 4, 5], target = 3.7, and k = 2, the two closest values would be [3, 4] since they have the smallest absolute differences from 3.7 (distances of 0.7 and 0.3 respectively).

The solution uses an in-order traversal of the BST (which visits nodes in ascending order) combined with a deque (double-ended queue) to maintain a sliding window of the k closest values. As it traverses:

• It first fills the deque with the first k values
• For subsequent values, it compares the distance of the new value with the distance of the leftmost value in the deque
• If the new value is closer, it removes the leftmost value and adds the new value
• The traversal can terminate early when values become too far from the target

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: A binary search tree is a special type of graph where each node has at most two children (left and right). The tree structure represents relationships between nodes through parent-child connections.

Is it a tree?

• Yes: The problem explicitly states we're working with a Binary Search Tree (BST), which is a hierarchical tree data structure.

DFS

• The flowchart leads us directly to DFS (Depth-First Search) when we confirm it's a tree structure.

Conclusion: The flowchart suggests using a DFS approach for this problem.

This makes perfect sense for "Closest Binary Search Tree Value II" because:

1. We need to traverse the entire tree (or most of it) to examine all node values
2. DFS allows us to leverage the BST property - we can perform an in-order traversal which visits nodes in sorted order
3. The sorted order from in-order DFS traversal helps us maintain a sliding window of the k closest values efficiently
4. We can potentially optimize by terminating the traversal early once we know remaining values will be farther from the target

The solution indeed uses DFS through recursive in-order traversal, visiting left subtree first, then processing the current node, and finally the right subtree. This pattern is the classic DFS approach for tree problems.

Intuition

The key insight comes from understanding two important properties: the BST structure and the nature of finding "k closest" values.

In a BST, an in-order traversal gives us values in sorted ascending order. This is crucial because when values are sorted, the closest values to any target must be contiguous in this sorted sequence. Think about it - if you have sorted values [1, 2, 3, 4, 5] and target 3.2, the closest values will form a continuous window around the target, not scattered randomly.

This leads us to a sliding window approach. As we traverse the tree in-order:

1. We maintain a window of size k containing our current best candidates
2. For each new value we encounter, we compare distances to decide if it should replace an existing value in our window

The clever part is using a deque (double-ended queue) to maintain this window. Why? Because as we traverse in ascending order:

• Early in the traversal, all values are smaller than or equal to the target
• Later in the traversal, values become larger than the target
• The leftmost value in our deque (the smallest) will always be the farthest from the target among our current candidates when we encounter larger values

This means we only need to compare the new value's distance with the leftmost value's distance. If the new value is closer, we slide the window by removing from the left and adding to the right.

An optimization emerges: once the incoming values become so large that they're farther from the target than our leftmost value, all subsequent values (being even larger in the sorted order) will also be too far. This allows us to terminate the traversal early rather than visiting the entire tree.

This approach elegantly combines the BST property with a sliding window technique to solve the problem in O(n) time with O(k) extra space for the deque.

Solution Approach

The implementation uses DFS with in-order traversal combined with a deque to maintain the k closest values:

Data Structures Used:

• deque: A double-ended queue to maintain a sliding window of k values
• Recursive call stack for DFS traversal

Algorithm Steps:

1. Initialize the deque: Create an empty deque q that will store up to k values.

2. Perform in-order DFS traversal: The dfs function recursively traverses the tree:

   dfs(root.left)  # Visit left subtree first
   # Process current node
   dfs(root.right) # Visit right subtree last

3. Process each node during traversal:

   • If deque has less than k elements: Simply append the current value

     if len(q) < k:
         q.append(root.val)

   • If deque is full (has k elements):

     • Compare the distance of the current value with the leftmost value in deque
     • Since we're traversing in ascending order, the leftmost value is the smallest
     • If abs(root.val - target) >= abs(q[0] - target), we can stop early because:
       • The current value is farther from target than our worst candidate
       • All subsequent values (being larger) will be even farther
     • Otherwise, slide the window:

       q.popleft()      # Remove the farthest value (leftmost)
       q.append(root.val)  # Add the closer value

4. Early termination optimization: When we encounter a value that's farther from the target than our worst candidate in the deque, we return immediately. This prevents unnecessary traversal of the remaining larger values in the tree.

5. Return the result: Convert the deque to a list and return.

Time Complexity: O(n) in worst case where we traverse all nodes, but often better due to early termination.

Space Complexity: O(k) for the deque plus O(h) for the recursion stack, where h is the height of the tree.

The elegance of this solution lies in leveraging the sorted nature of in-order traversal to maintain a sliding window efficiently, requiring only one comparison per node after the deque is full.

Example Walkthrough

Let's walk through finding the k=2 closest values to target=3.8 in this BST:

       4
      / \
     2   5
    / \
   1   3

Initial Setup:

• Target = 3.8
• k = 2
• deque q = []

In-order Traversal (visits nodes in ascending order: 1→2→3→4→5):

Step 1: Visit node 1

• q is empty (len < k), so append 1
• q = [1]

Step 2: Visit node 2

• q has 1 element (len < k), so append 2
• q = [1, 2]
• Now q is full with k=2 elements

Step 3: Visit node 3

• q is full, need to compare distances:
  • Distance of new value 3 from target: |3 - 3.8| = 0.8
  • Distance of leftmost value 1 from target: |1 - 3.8| = 2.8
  • Since 0.8 < 2.8, value 3 is closer than value 1
• Slide the window: remove 1 from left, add 3 to right
• q = [2, 3]

Step 4: Visit node 4

• q is full, compare distances:
  • Distance of new value 4 from target: |4 - 3.8| = 0.2
  • Distance of leftmost value 2 from target: |2 - 3.8| = 1.8
  • Since 0.2 < 1.8, value 4 is closer than value 2
• Slide the window: remove 2 from left, add 4 to right
• q = [3, 4]

Step 5: Visit node 5

• q is full, compare distances:
  • Distance of new value 5 from target: |5 - 3.8| = 1.2
  • Distance of leftmost value 3 from target: |3 - 3.8| = 0.8
  • Since 1.2 > 0.8, value 5 is NOT closer than value 3
• Early termination triggered! Since we're traversing in ascending order and value 5 is already farther than our worst candidate (3), all subsequent values would be even farther.
• Return immediately with q = [3, 4]

Result: [3, 4] are the 2 closest values to 3.8

The key insight is that the deque maintains a sliding window of the k closest values seen so far, and the sorted nature of in-order traversal allows us to stop early once values become too far from the target.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the number of nodes in the binary tree.

The algorithm performs an in-order traversal of the BST using DFS. In the worst case, it visits every node in the tree exactly once. Even though there's an early termination condition (if abs(root.val - target) >= abs(q[0] - target): return), this optimization only helps when we've found k values and the remaining nodes are farther from the target. In the worst case scenario where all nodes are very close to the target or when the target is at one extreme of the tree's value range, we still need to traverse all n nodes.

Space Complexity: O(h + k) where h is the height of the tree and k is the number of closest values to find.

The space complexity consists of two components:

• The recursive call stack for DFS traversal uses O(h) space, where h is the height of the tree. In the worst case of a skewed tree, h = n, making it O(n). In the best case of a balanced tree, h = log(n).
• The deque q stores at most k elements at any time, contributing O(k) space.
• The final list conversion also takes O(k) space, but this doesn't add to the asymptotic complexity.

Therefore, the total space complexity is O(h + k).

Common Pitfalls

1. Incorrect Early Termination Logic

The Pitfall: A common mistake is using the wrong comparison for early termination. Some might incorrectly write:

if abs(node.val - target) > abs(closest_values_queue[0] - target):
    return

Instead of using >=. This subtle difference can cause the algorithm to miss valid candidates when the distances are equal.

Why It's Wrong: When distances are equal, we should still terminate because:

• We already have k values with distances less than or equal to the current distance
• Since we're traversing in ascending order, all subsequent values will have distances greater than or equal to the current one
• Missing the equal case means unnecessary traversal continues

The Fix: Always use >= in the comparison to handle the edge case where distances are equal:

if abs(node.val - target) >= abs(closest_values_queue[0] - target):
    return

2. Assuming Target is an Integer

The Pitfall: Treating the target as an integer when it's actually a float, which can lead to incorrect distance calculations:

# Wrong: assuming target is int
def closestKValues(self, root: Optional[TreeNode], target: int, k: int) -> List[int]:

Why It's Wrong: The target can be any floating-point number (e.g., 3.7, 2.5), and treating it as an integer would either cause type errors or incorrect results due to implicit conversion.

The Fix: Always declare target as float and handle floating-point arithmetic properly:

def closestKValues(self, root: Optional[TreeNode], target: float, k: int) -> List[int]:

3. Not Handling the Case When k = 0

The Pitfall: The code doesn't explicitly handle when k = 0, which should return an empty list:

# Current code will work but isn't explicit about this edge case
if len(closest_values_queue) < k:  # When k=0, this never executes
    closest_values_queue.append(node.val)

Why It Matters: While the current implementation technically handles k=0 correctly (returns empty list), it's not immediately obvious and could confuse maintainers.

The Fix: Add an explicit check at the beginning:

def closestKValues(self, root: Optional[TreeNode], target: float, k: int) -> List[int]:
    if k == 0 or root is None:
        return []
  
    # Rest of the implementation...

4. Misunderstanding the Sliding Window Direction

The Pitfall: Some might think they should remove from the right side of the deque instead of the left:

# Wrong approach:
closest_values_queue.pop()  # Removes from right
closest_values_queue.append(node.val)

Why It's Wrong: In in-order traversal, we visit nodes in ascending order. The leftmost element in the deque is the smallest value, and as we encounter larger values, we need to check if they're closer to the target than the smallest (leftmost) value. Removing from the right would break this logic.

The Fix: Always remove from the left when the deque is full:

closest_values_queue.popleft()  # Correct: removes the smallest/leftmost value
closest_values_queue.append(node.val)