题目描述：

LeetCode 783. Minimum Distance Between BST Nodes

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:

1. The size of the BST will be between 2 and 100.
2. The BST is always valid, each node's value is an integer, and each node's value is different.

题目大意：

给定二叉搜索树（BST）根节点为root，求任意两节点的最小差值。

解题思路：

递归（Recursion)

Python代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def minDiffInBST(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        ans = 0x7FFFFFFF
        if root.left:
            lmax = root.left
            while lmax.right: lmax = lmax.right
            ans = min(ans, root.val - lmax.val)
            ans = min(ans, self.minDiffInBST(root.left))
        if root.right:
            rmin = root.right
            while rmin.left: rmin = rmin.left
            ans = min(ans, rmin.val - root.val)
            ans = min(ans, self.minDiffInBST(root.right))
        return ans

 

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