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
"""
# traverse node from min to max
self.last_node = None
self.min_diff = float('inf') def dfs(node):
if not node: return
dfs(node.left)
if self.last_node:
self.min_diff = min(self.min_diff, node.val-self.last_node.val)
self.last_node = node
dfs(node.right) dfs(root)
return self.min_diff
# 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
"""
# traverse node from min to max
last_node = None
ans = float("inf")
node = root
stack = []
while node:
stack.append(node)
node = node.left
while stack:
node = stack.pop()
if last_node:
ans = min(ans, node.val-last_node.val)
last_node = node
node = node.right
while node:
stack.append(node)
node = node.left
return ans

Morris Traversal:

# 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
"""
# traverse node from min to max
last_node = None
ans = float("inf") cur = root
while cur:
if cur.left is None:
if last_node:
ans = min(ans, cur.val-last_node.val)
last_node = cur
cur = cur.right
else:
tmp = cur.left
while tmp.right and not (tmp.right is cur):
tmp = tmp.right
if not tmp.right:
tmp.right = cur
cur = cur.left
else:
tmp.right = None
if last_node:
ans = min(ans, cur.val-last_node.val)
last_node = cur
cur = cur.right
return ans

leetcode 783. Minimum Distance Between BST Nodes 以及同样的题目 530. Minimum Absolute Difference in BST的更多相关文章

  1. 51. leetcode 530. Minimum Absolute Difference in BST

    530. Minimum Absolute Difference in BST Given a binary search tree with non-negative values, find th ...

  2. 【leetcode_easy】530. Minimum Absolute Difference in BST

    problem 530. Minimum Absolute Difference in BST 参考 1. Leetcode_easy_530. Minimum Absolute Difference ...

  3. LeetCode 530. Minimum Absolute Difference in BST (二叉搜索树中最小绝对差)

    Given a binary search tree with non-negative values, find the minimum absolute difference between va ...

  4. [LeetCode] Minimum Absolute Difference in BST 二叉搜索树的最小绝对差

    Given a binary search tree with non-negative values, find the minimum absolute difference between va ...

  5. LeetCode Minimum Absolute Difference in BST

    原题链接在这里:https://leetcode.com/problems/minimum-absolute-difference-in-bst/#/description 题目: Given a b ...

  6. 【LeetCode】530. Minimum Absolute Difference in BST 解题报告(Java & Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 Java解法 Python解法 日期 题目地址:ht ...

  7. [LeetCode&Python] Problem 530. Minimum Absolute Difference in BST

    Given a binary search tree with non-negative values, find the minimum absolute difference between va ...

  8. C#LeetCode刷题之#530-二叉搜索树的最小绝对差(Minimum Absolute Difference in BST)

    问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/4123 访问. 给定一个所有节点为非负值的二叉搜索树,求树中任意两 ...

  9. [Swift]LeetCode530. 二叉搜索树的最小绝对差 | Minimum Absolute Difference in BST

    Given a binary search tree with non-negative values, find the minimum absolute difference between va ...

随机推荐

  1. CDOJ 1220 The Battle of Guandu

    The Battle of Guandu Time Limit: 6000/3000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Oth ...

  2. XV内存变化

  3. Problem 2125 简单的等式(FZU),,数学题。。。

    Problem 2125 简单的等式 Time Limit: 1000 mSec Memory Limit : 32768 KB  Problem Description 现在有一个等式如下:x^2+ ...

  4. JS Enter键跳转 控件获得焦点

    //回车跳转 jQuery(document).ready(function () { //$(':input:text:first').focus(); jQuery(':input:enabled ...

  5. 624. Maximum Distance in Arrays

    Problem statement Given m arrays, and each array is sorted in ascending order. Now you can pick up t ...

  6. 关于datetime,date,timestamp,year,time时间类型小结

    关于datetime,date,timestamp,year,time时间类型 datetime占用8个字节 日期范围:”1000-01-01 00:00:00” 到”9999-12-31 23:59 ...

  7. loadrunner 并发操作集合点配置

    在loadrunner的虚拟用户中,术语concurrent(并发)和simultaneous(同时)存在一些区别,concurrent 是指虚拟场景中参于运行的虚拟用户.而simultaneous与 ...

  8. django学习之- json序列化

    序列化操作 - Errordict - 自定义Encoder - django的模块可以直接序列化 第一种: from django.core import serializers # 通过这个模块对 ...

  9. 没啥用,更换注册表信息使webbrower选择适合的版本

    /// <summary>            /// 修改注册表信息来兼容当前程序            ///             /// </summary>   ...

  10. 洛谷——P2916 [USACO08NOV]为母牛欢呼Cheering up the Cows

    https://www.luogu.org/problem/show?pid=2916 题目描述 Farmer John has grown so lazy that he no longer wan ...