Minimum Absolute Difference in BST

Difficulty: Easy | Acceptance: 59.40% | Paid: No Topics: Tree, Depth-First Search, Breadth-First Search, Binary Search Tree, Binary Tree

Given the root of a Binary Search Tree (BST), return the minimum absolute difference between the values of any two different nodes in the tree.

• Examples
• Constraints
• Approach 1: In-order Traversal (Iterative)
• Approach 2: In-order Traversal (Recursive)
• Approach 3: Flatten and Compare

Examples

Input: root = [4,2,6,1,3]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.
The given tree [4,2,6,1,3] is represented by the following diagram:
      4
    /   \
  2      6
 / \    
1   3  
while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.

Input: root = [1,0,48,null,null,12,49]
Output: 1
Explanation:
The minimum difference occurs between node 48 and node 12, and also between node 48 and node 49.

Constraints

- The number of nodes in the tree is in the range [2, 10^4].
- 0 <= Node.val <= 10^5

Approach 1: In-order Traversal (Iterative)

Intuition In-order traversal of a Binary Search Tree visits nodes in strictly ascending order. Therefore, the minimum absolute difference must exist between two adjacent nodes visited during this traversal.

Steps

• Initialize a stack to simulate the recursion and variables to track the previous node and the minimum difference found so far.
• While the stack is not empty or the current node is not null:
  • Traverse to the leftmost node, pushing nodes onto the stack.
  • Pop a node from the stack (this is the current node to process).
  • If a previous node exists, calculate the difference between the current node’s value and the previous node’s value. Update the minimum difference if this new difference is smaller.
  • Set the current node as the previous node for the next iteration.
  • Move to the right child of the current node.
• Return the minimum difference found.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def getMinimumDifference(self, root: Optional[TreeNode]) -> int:
        stack = []
        prev = None
        min_diff = float('inf')
        curr = root
        
        while stack or curr:
            while curr:
                stack.append(curr)
                curr = curr.left
            
            curr = stack.pop()
            
            if prev is not None:
                min_diff = min(min_diff, curr.val - prev.val)
            
            prev = curr
            curr = curr.right
            
        return min_diff

Complexity

• Time: O(N) — We visit each node exactly once.
• Space: O(H) — The stack stores at most the height of the tree (H). In the worst case (skewed tree), this is O(N).
• Notes: This is the most space-efficient approach for iterative solutions as it avoids storing all node values.

Approach 2: In-order Traversal (Recursive)

Intuition Similar to the iterative approach, we leverage the property of BST that in-order traversal yields sorted values. We use recursion to traverse the tree and maintain the previous node’s value and the minimum difference using class member variables or closures.

Steps

• Initialize member variables (or closure variables) to store the previous node value and the minimum difference.
• Define a recursive helper function to perform in-order traversal.
• In the helper function:
  • Recursively call the function on the left child.
  • Process the current node: if a previous node exists, update the minimum difference. Update the previous node to the current node.
  • Recursively call the function on the right child.
• Start the recursion from the root and return the minimum difference.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def getMinimumDifference(self, root: Optional[TreeNode]) -> int:
        self.prev = None
        self.min_diff = float('inf')
        
        def inorder(node):
            if not node:
                return
            
            inorder(node.left)
            
            if self.prev is not None:
                self.min_diff = min(self.min_diff, node.val - self.prev.val)
            self.prev = node
            
            inorder(node.right)
            
        inorder(root)
        return self.min_diff

Complexity

• Time: O(N) — We visit each node exactly once.
• Space: O(H) — The recursion stack uses space proportional to the height of the tree.
• Notes: This approach is often cleaner to write than the iterative version but uses the call stack.

Approach 3: Flatten and Compare

Intuition We can perform an in-order traversal to collect all node values into a list. Since the list will be sorted, we can then iterate through the list once to find the minimum difference between adjacent elements.

Steps

• Initialize an empty list to store values.
• Perform an in-order traversal (iterative or recursive) of the BST, appending each node’s value to the list.
• Iterate through the list from index 1 to the end.
• For each element, calculate the difference with the previous element and update the global minimum difference.
• Return the minimum difference.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def getMinimumDifference(self, root: Optional[TreeNode]) -> int:
        values = []
        
        def inorder(node):
            if not node:
                return
            inorder(node.left)
            values.append(node.val)
            inorder(node.right)
            
        inorder(root)
        
        min_diff = float('inf')
        for i in range(1, len(values)):
            min_diff = min(min_diff, values[i] - values[i-1])
            
        return min_diff

Complexity

• Time: O(N) — Traversal takes O(N) and iterating through the list takes O(N).
• Space: O(N) — We store all N node values in the list.
• Notes: This approach is conceptually simple but uses more memory than the previous approaches because it stores the entire sequence of values.