leetcode 783. Minimum Distance Between BST Nodes 以及同样的题目 530. Minimum Absolute Difference in BST
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:
- The size of the BST will be between 2 and
100. - 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
- 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 ...
- 【leetcode_easy】530. Minimum Absolute Difference in BST
problem 530. Minimum Absolute Difference in BST 参考 1. Leetcode_easy_530. Minimum Absolute Difference ...
- 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 ...
- [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 ...
- 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 ...
- 【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 ...
- [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 ...
- 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 ...
- [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
- CDOJ 1220 The Battle of Guandu
The Battle of Guandu Time Limit: 6000/3000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Oth ...
- XV内存变化
- 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+ ...
- 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 ...
- 624. Maximum Distance in Arrays
Problem statement Given m arrays, and each array is sorted in ascending order. Now you can pick up t ...
- 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 ...
- 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 ...
- 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 ...
- 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> ...
- 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 ...