Given a Binary Search Tree (BST) with the root noderoot, 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:

1. The size of the BST will be between 2 and 100 .
2. 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.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    public int minDiffInBST(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        dfs(root, list);

        int ans = Integer.MAX_VALUE;
        for (int i = 1; i < list.size(); i++){
            ans = Math.min(ans, list.get(i) - list.get(i - 1));
        }

        return ans;
    }

    private void dfs(TreeNode root, List<Integer> list){
        if (root == null) return;

        dfs(root.left, list);
        list.add(root.val);
        dfs(root.right, list);
    }
}