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.

思路：
BST + 中序，再求minimum difference

/**
 * 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) {
        vis = new ArrayList<Integer>();
        dfs(root);
        int min = 0x3f3f3f3f;
        for (int i = 1; i < vis.size(); ++i) {
            int diff = vis.get(i) - vis.get(i - 1);
            min = Math.min(min, diff);
        }
        return min;
    }

    List<Integer> vis;
    public void dfs(TreeNode root) {
        if (root == null) return;
        dfs(root.left);  
        vis.add(root.val);
        dfs(root.right);
    }
}