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

题目地址：https://leetcode.com/problems/minimum-distance-between-bst-nodes/description/

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、 ThesizeoftheBSTwillbebetween2and100.；
2、 TheBSTisalwaysvalid,eachnode'svalueisaninteger,andeachnode'svalueisdifferent.；

题目大意

求BST的两个节点之间的最小差值。

解题方法

中序遍历

看见BST想中序遍历是有序的啊~所以先进性中序遍历，得到有序列表，然后找出相邻的两个节点差值的最小值即可。

# 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)])

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二刷的时候注意到和530. Minimum Absolute Difference in BSTopen in new window是完全一样的题，果然同样的代码就直接通过了。。不懂这个题的意义是什么。。

# 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)

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