[LeetCode&Python] Problem 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(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
        """
        diff=100000
        last_visited=None
        stack=[]
        node=root
        while node or stack:
            if node:
                stack.append(node)
                node=node.left
            else:
                node=stack.pop()
                if last_visited and abs(last_visited.val-node.val)<diff:
                    diff=abs(last_visited.val-node.val)
                last_visited=node
                node=node.right
        return diff

原文地址：https://www.cnblogs.com/chiyeung/p/10011051.html