【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/
题目地址:https://leetcode.com/problems/minimum-absolute-difference-in-bst/#/description
题目描述
Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.
Example:
Input:
1
\
3
/
2
Output:
1
Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).
Note: There are at least two nodes in this BST.
题目大意
判断一个BST中任意两个节点之间的差的绝对值的最小值。
解题方法
Java解法
找出BST中两个节点的最小差距值。
第一遍没有思路,第二次看就想到了中序遍历,BST的中序遍历是有序。应该可以通过数组保存的形式,但是看了别人的做法,发现直接用个外部的变量就能保存最小的值。另外还要一个prev保存上一个节点。
为什么是当前的值-上一个节点的值呢?因为我们的遍历是有序的,所以当前节点比前一个节点大,这样相减就可以保证结果是正的。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int minDiff = Integer.MAX_VALUE;
TreeNode prev = null;
public int getMinimumDifference(TreeNode root) {
inOrder(root);
return minDiff;
}
public void inOrder(TreeNode root){
if(root == null){
return;
}
inOrder(root.left);
if(prev !=null) minDiff= Math.min(minDiff, root.val - prev.val);
prev = root;
inOrder(root.right);
}
}
Python解法
二刷,python。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def getMinimumDifference(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
2017 年 4 月 8 日
2018 年 11 月 14 日 —— 很严重的雾霾
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