Problem Description

You are given the root of a binary search tree (BST) and a target value (which can be a floating-point number). Your task is to find the value in the BST that is closest to the target.

The "closest" value means the node value that has the minimum absolute difference with the target. In other words, you need to find a node value v such that |v - target| is minimized.

If there are multiple node values that have the same minimum distance to the target (a tie situation), you should return the smallest value among them.

For example:

• If the BST contains values [4, 2, 5, 1, 3] and the target is 3.714286, the closest value would be 4 since |4 - 3.714286| = 0.285714 is the smallest difference.
• If the target is exactly between two values, like 3.5 with nodes 3 and 4 both having a distance of 0.5, you would return 3 (the smaller value).

The solution leverages the BST property where left subtree values are smaller than the root, and right subtree values are larger. This allows for efficient traversal by comparing the target with the current node value to decide whether to go left or right, while keeping track of the closest value found so far.

Flowchart Walkthrough

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

Is it a graph?

• Yes: While this is a binary search tree problem, a tree is a special type of graph (specifically, a connected acyclic graph). The BST has nodes connected by edges (parent-child relationships).

Is it a tree?

• Yes: The problem explicitly states we're working with a binary search tree, which is a tree data structure where each node has at most two children.

DFS

• Since we answered "yes" to the tree question, the flowchart leads us directly to DFS (Depth-First Search).

Conclusion: The flowchart suggests using DFS for this problem.

This makes sense because:

1. We need to traverse the tree to examine node values and compare them with the target
2. DFS allows us to efficiently navigate the BST by leveraging its ordering property (left subtree has smaller values, right subtree has larger values)
3. We can recursively explore either the left or right subtree based on how the current node's value compares to the target, effectively pruning unnecessary branches
4. During the traversal, we maintain and update the closest value found so far

The DFS approach in the solution recursively visits nodes, updates the answer when a closer value is found, and intelligently chooses whether to explore the left or right subtree based on the BST property.

Intuition

The key insight comes from understanding the binary search tree property: for any node, all values in its left subtree are smaller, and all values in its right subtree are larger. This ordered structure allows us to make intelligent decisions about which direction to search.

Think of it like searching for a house number on a street where houses are numbered in order. If you're looking for house #50 and you're currently at house #30, you know you need to go forward (right). If you're at house #70, you need to go backward (left).

When we're at any node in the BST, we can ask: "Is the target value smaller or larger than the current node's value?" This comparison tells us which subtree potentially contains values closer to our target:

• If target < node.val, closer values might exist in the left subtree (smaller values)
• If target > node.val, closer values might exist in the right subtree (larger values)

However, we can't immediately discard the current node! Even if the target is smaller than the current value, the current node might still be the closest. For example, if we have a node with value 10 and the target is 9.9, we should explore the left subtree, but 10 might end up being the closest value if no nodes exist between 9.9 and 10.

This leads to our strategy:

1. At each node, calculate the absolute difference |node.val - target|
2. Keep track of the node with the minimum difference seen so far
3. If there's a tie (same difference), choose the smaller value
4. Continue searching in the direction where closer values might exist
5. The recursion naturally handles visiting all potentially relevant nodes

The beauty of this approach is that we don't need to visit every node in the tree. The BST property allows us to prune entire subtrees that can't possibly contain the answer, making the algorithm efficient with O(h) time complexity where h is the height of the tree.

Solution Approach

The solution implements a recursive DFS approach to traverse the binary search tree and find the closest value to the target.

Implementation Details:

We define a recursive helper function dfs(node) that performs the traversal. The function maintains two variables in the outer scope:

• ans: stores the node value that is currently closest to the target
• diff: stores the minimum absolute difference found so far (initialized to infinity)

Step-by-step Algorithm:

1. Base Case: If the current node is None, we return immediately (end of path reached).

2. Calculate Distance: For the current node, we calculate the absolute difference between its value and the target:

   nxt = abs(target - node.val)

3. Update Answer: We update our answer in two cases:

   • If the current difference nxt is smaller than the best difference diff we've seen
   • If the difference is the same but the current node's value is smaller (handling ties)

   if nxt < diff or (nxt == diff and node.val < ans):
       diff = nxt
       ans = node.val

4. Choose Direction: Based on the BST property, we decide which subtree to explore:

   node = node.left if target < node.val else node.right

   If the target is less than the current node's value, we go left (to find smaller values). Otherwise, we go right (to find larger values).

5. Recursive Call: We continue the search in the chosen direction:

   dfs(node)

Why This Works:

The algorithm leverages the BST property to avoid unnecessary traversals. While we might miss the closest value if we only went in one direction, by checking each node along our path and updating the answer before choosing a direction, we ensure we don't miss any potential candidates.

For example, if we're at node with value 5 and the target is 4.6, we'll:

• Record 5 as a candidate (difference = 0.4)
• Move left to find values smaller than 5
• If we find 4 (difference = 0.6), we keep 5 as the answer
• If we find 4.5 (difference = 0.1), we update to 4.5

The time complexity is O(h) where h is the height of the tree, as we follow a single path from root to leaf. The space complexity is also O(h) due to the recursion stack.

Example Walkthrough

Let's walk through finding the closest value to target 4.2 in this BST:

       5
      / \
     3   7
    / \   \
   1   4   9

Step 1: Start at root (value = 5)

• Calculate difference: |5 - 4.2| = 0.8
• Update: ans = 5, diff = 0.8
• Since 4.2 < 5, go left to node 3

Step 2: Visit node 3

• Calculate difference: |3 - 4.2| = 1.2
• Compare: 1.2 > 0.8, so don't update answer
• Since 4.2 > 3, go right to node 4

Step 3: Visit node 4

• Calculate difference: |4 - 4.2| = 0.2
• Compare: 0.2 < 0.8, so update answer
• Update: ans = 4, diff = 0.2
• Since 4.2 > 4, we would go right, but right child is None

Step 4: End of path

• Return with final answer: 4

The algorithm found the closest value 4 with a difference of 0.2 from the target 4.2.

Example with a tie: If the target was 4.5 instead:

• Node 5: difference = 0.5
• Node 4: difference = 0.5 (same difference)
• Since both have the same difference but 4 < 5, we return 4

Notice how we only visited 3 nodes instead of all 6 nodes in the tree, demonstrating the efficiency of using the BST property to guide our search.

Solution Implementation

Time and Space Complexity

The time complexity is O(h) where h is the height of the binary search tree. In the worst case when the tree is completely unbalanced (essentially a linked list), h = n, making the time complexity O(n). In the best case when the tree is perfectly balanced, h = log(n), making the time complexity O(log n). The algorithm leverages the BST property by choosing to traverse either left or right based on the comparison between target and current node value, visiting at most one node per level.

The space complexity is O(h) due to the recursive call stack, where h is the height of the tree. In the worst case of an unbalanced tree, this becomes O(n). In the best case of a balanced tree, this is O(log n). The recursive DFS function creates a new stack frame for each level of recursion, and the maximum depth of recursion equals the height of the tree.

Note: The reference answer states both complexities as O(n), which represents the worst-case scenario when the BST is completely unbalanced.

Common Pitfalls

1. Not Handling Tie-Breaking Correctly

The Pitfall: When two node values have the same distance from the target, forgetting to return the smaller value can lead to incorrect results. For instance, if target is 2.5 and the BST contains both 2 and 3, both have a distance of 0.5. The problem requires returning 2 (the smaller value), but a naive implementation might return whichever was encountered last.

Incorrect Implementation:

# Wrong: Only checks if difference is smaller, ignores ties
if current_difference < min_difference:
    min_difference = current_difference
    closest_value = node.val

Correct Solution:

# Correct: Handles both smaller difference AND tie-breaking
if current_difference < min_difference or \
   (current_difference == min_difference and node.val < closest_value):
    min_difference = current_difference
    closest_value = node.val

2. Early Termination When Finding Exact Match

The Pitfall: Some might optimize by immediately returning when node.val == target, thinking they've found the perfect match. However, this prevents proper tie-breaking when the target is exactly between two values.

Incorrect Implementation:

def dfs(node):
    if node is None:
        return
  
    # Wrong: Early return prevents checking other equal-distance nodes
    if node.val == target:
        return node.val
  
    # ... rest of logic

Why It's Wrong: If target is 2.5 and we encounter node 2 first, then node 3, we might incorrectly return 3 if we store it as "exact match" without comparing which is smaller.

3. Traversing Both Subtrees Instead of One

The Pitfall: Traversing the entire tree defeats the purpose of using a BST's properties for optimization. This changes complexity from O(h) to O(n).

Incorrect Implementation:

def dfs(node):
    if node is None:
        return
  
    # Update closest value logic...
  
    # Wrong: Explores both subtrees unnecessarily
    dfs(node.left)
    dfs(node.right)

Correct Approach: The solution should choose only one direction based on the comparison between target and current node value, ensuring O(h) complexity while still checking all nodes along the chosen path.

4. Integer Division or Rounding Issues

The Pitfall: Using integer operations or rounding when dealing with the float target can cause precision loss.

Incorrect Implementation:

# Wrong: Converting target to int loses precision
target_int = int(target)
current_difference = abs(target_int - node.val)

Correct Solution: Always maintain the target as a float and use abs(target - node.val) to preserve precision in distance calculations.