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.

# 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):
        """
        :type root: TreeNode
        :rtype: int
        """
        a = []
        def inorder(root):
            if root.left:
                inorder(root.left)
            a.append(root.val)
            if root.right:
                inorder(root.right)
        res = 100000
        inorder(root)
        for i in range(len(a)-1):
            res = min(res,a[i+1]-a[i])
        return res

原文地址：https://www.cnblogs.com/bernieloveslife/p/9836257.html