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.

这道题可以利用一下BST的性质，最小距离要么是一个树的root和左子树中最大值的差，要么是右子树最小值和root的差，要么来自左右子树的最小距离。可以通过inorder traversal 来解决

/**
 * 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) {
        if(root == null)
            return 0;

        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        while(p != null) {
            stack.push(p);
            p = p.left;
        }

        int res = Integer.MAX_VALUE;
        Integer last = null;

        while(stack.size() > 0) {
            TreeNode temp = stack.pop();
            if(last == null) {
                last = temp.val;
            } else {
                res = Math.min(res, temp.val - last);
                last = temp.val;
            }

            if(temp.right != null) {
                p = temp.right;
                while(p != null) {
                    stack.push(p);
                    p = p.left;
                }
            }
        }

        return res;
    }
}