530. Minimum Absolute Difference in BST

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.

My solution:
// didn't make use of the characteristics of BST. Anyway, it got accepted.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int getMinimumDifference(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        add(list,root);
        Collections.sort(list);
        ArrayList<Integer> list2 = new ArrayList<Integer>();
        for(int i = 0; i < list.size() - 1; i++) {
            int distance = list.get(i + 1) - list.get(i);
            list2.add(distance);
        }
        
        Collections.sort(list2);
        
        return list2.get(0);
    }
    
    public void add(ArrayList<Integer> list, TreeNode root){
        if(root == null) return;
        list.add(root.val);
        
        add(list, root.left);
        add(list, root.right);
    }
}

Solution from Leetcode:

Since this is a BST, the inorder traversal of its nodes results in a sorted list of values. Thus, the minimum absolute difference must occur in any adjacently traversed nodes. I use the global variable "prev" to keep track of each node's inorder predecessor.

public class Solution {
    
    int minDiff = Integer.MAX_VALUE;
    TreeNode prev;
    
    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);
    }
}