530. Minimum Absolute Difference in BST¶
Edited by Charles
Easy · Python · Binary Tree
Problem
Read the original problem on LeetCode.
The official page contains the problem statement, examples, and constraints.
Explanation, examples & other approaches
Core idea
Iteration:
Reference to 98. Validate Binary Search Tree for the core idea in an order traversal.
Original study notes follow, including intermediate snippets and earlier attempts.
Iteration:
Reference to 98. Validate Binary Search Tree for the core idea in an order traversal.
# 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:
minDiff = float('inf')
stack = []
cur = root
pre = None
while cur or stack:
if cur:
stack.append(cur)
cur = cur.left
else:
cur = stack.pop()
if pre:
minDiff = min(minDiff, cur.val - pre.val)
pre = cur
cur = cur.right
return minDiff
Recursion:
Still the basic idea of the inorder traversal but in the recursive function.
Study record
| Field | Value |
|---|---|
| Attempts | 1 |
| Mastery | ❚ |
| The Latest Attempt | January 16, 2026 |
Python solution¶
# 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 __init__(self):
self.minDiff = float('inf')
self.pre = None
def getMinimumDifference(self, root: Optional[TreeNode]) -> int:
self.traverse(root)
return self.minDiff
def traverse(self, node):
if not node:
return
self.traverse(node.left)
if self.pre:
self.minDiff = min(self.minDiff, node.val - self.pre.val)
self.pre = node
self.traverse(node.right)