【LeetCode】783. Minimum Distance Between BST Nodes 解题报告(Python)
2024-09-05 03:47:08
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
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:
- 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.
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)])
二刷的时候注意到和530. Minimum Absolute Difference in BST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)
日期
2018 年 2 月 28 日
2018 年 11 月 14 日 —— 很严重的雾霾
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