1. 力扣530:二叉搜索树的最小绝对差
1.1 题目:
给你一个二叉搜索树的根节点 root
,返回 树中任意两不同节点值之间的最小差值 。
差值是一个正数,其数值等于两值之差的绝对值。
示例 1:
输入:root = [4,2,6,1,3] 输出:1
示例 2:
输入:root = [1,0,48,null,null,12,49] 输出:1
提示:
- 树中节点的数目范围是
[2, 104]
0 <= Node.val <= 105
1.2 思路
题目虽然说求任意两个不同的节点的差值,但很容易发现,排序以后,只需要比较两个相邻节点的差值即可。
1.3 题解:
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {// 用列表记录二叉树节点的值List<Integer> list = new ArrayList<>();public int getMinimumDifference(TreeNode root) {recursion(root);// 调用库方法排序Collections.sort(list);// 因为由题,该二叉树最少两个节点int targrt = list.get(0) - list.get(1);if(targrt < 0) {targrt = -targrt;}// 排序以后只需要比较两个相邻的元素的差值for(int i = 0; i < list.size() - 1; i++) {int diff = list.get(i) - list.get(i + 1);if(diff < 0) {diff = -diff;}targrt = targrt > diff ? diff : targrt;}return targrt;}private void recursion(TreeNode node) {if(node == null) {return;}list.add(node.val);recursion(node.left);recursion(node.right);}
}