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

@TOC

## # 题目描述

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.

## # 解题方法

### # 中序遍历

``````# 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
"""
vals = []
def inOrder(root):
if not root:
return
inOrder(root.left)
vals.append(root.val)
inOrder(root.right)
inOrder(root)
return min([vals[i + 1] - vals[i] for i in xrange(len(vals) - 1)])
``````

``````# 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
"""
self.res = float("inf")
self.prev = None
self.inOrder(root)
return self.res

def inOrder(self, root):
if not root: return
self.inOrder(root.left)
if self.prev:
self.res = min(self.res, root.val - self.prev.val)
self.prev = root
self.inOrder(root.right)
``````

## # 日期

2018 年 2 月 28 日 2018 年 11 月 14 日 —— 很严重的雾霾