783. Minimum Distance Between BST Nodes-二叉搜索树节点最小距离

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

Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/

Example 1:

Input: root = [4,2,6,1,3]
Output: 1

Example 2:

Input: root = [1,0,48,null,null,12,49]
Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 100].
• 0 <= Node.val <= 10^5

Solution in python:

# 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 minDiffInBST(self, root: TreeNode) -> int:
        prior = float('-inf')
        min_diff = float('inf')
        def midOrder(root):
            if root == None:
                return 
            else:
                midOrder(root.left)
                nonlocal prior
                diff = root.val - prior
                prior = root.val
                nonlocal min_diff
                if diff < min_diff:
                    min_diff = diff
                midOrder(root.right)
        midOrder(root)
        return min_diff