530. Minimum Absolute Difference in BST == 783. Minimum Distance Between BST Nodes

530. Minimum Absolute Difference in BST == 783. Minimum Distance Between BST Nodes

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:

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.

My Solutions:

class Solution {
    
    TreeNode pre;
    int res = Integer.MAX_VALUE;
    
    public int getMinimumDifference(TreeNode root) {
        inorder(root);
        return res;        
    }
    
    public void inorder(TreeNode root) {
        if (root == null) return;
        inorder(root.left);
        if (pre != null) res = Math.min(res, root.val - pre.val);
        pre = root;
        inorder(root.right);
    }
}

class Solution {
    
    int res = Integer.MAX_VALUE;
    Integer pre = null;
    
    public int minDiffInBST(TreeNode root) {
        if (root.left != null) minDiffInBST(root.left);
        if (pre != null) res = Math.min(res, root.val - pre);
        pre = root.val;
        if (root.right != null) minDiffInBST(root.right);
        return res;
    }
}