首页 > 其他 > 详细

783. Minimum Distance Between BST Nodes BST节点之间的最小距离

时间:2018-02-12 23:33:41      阅读:411      评论:0      收藏:0      [点我收藏+]

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.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
# 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
        """
        res = float(‘inf‘)
        pre = -float(‘inf‘)
 
        def find(root):
            nonlocal res
            nonlocal pre
            if root.left:
                find(root.left)
            res = min(res, root.val - pre)
            pre = root.val
            if root.right:
                find(root.right)
 
        find(root)
        return res







783. Minimum Distance Between BST Nodes BST节点之间的最小距离

原文:https://www.cnblogs.com/xiejunzhao/p/8445817.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!