Problem description
Given a Binary Search Tree (BST) with the root node root, return the minimum difference between the values of any two different nodes in the tree.
Example :
Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.
The given tree [4,2,6,1,3,null,null] 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. Note:
The size of the BST will be between 2 and 100. The BST is always valid, each node’s value is an integer, and each node’s value is different.
Solution
- We can do inorder traversal and keep record of current minimum difference and current node’s predecessor’s value
Below is python implementation
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def minDiffInBST(self, root: TreeNode) -> int:
self.prev = -float('inf')
self.ans = float('inf')
def inorder(root:TreeNode)->None:
if not root:
return
inorder(root.left)
self.ans = min(self.ans, root.val - self.prev)
self.prev = root.val
inorder(root.right)
return
inorder(root)
return self.ans