# Minimum Distance (Difference) Between BST Nodes

`Binary Search Tree`

**Almost** **same as** [**Minimum Absolute Difference in BST**](https://leetcode.com/problems/minimum-absolute-difference-in-bst/)

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.

## Solution

**In-order Traversal of BST** output the node value in sorted order; using a global variable prev to store previous value for comparison.

[https://leetcode.com/problems/minimum-distance-between-bst-nodes/solution/](https://leetcode.com/problems/minimum-distance-between-bst-nodes/solution/)

### Approach #1: Write to Array <a href="#approach-1-write-to-array-accepted" id="approach-1-write-to-array-accepted"></a>

**Complexity Analysis**

* Time Complexity: O(N), where N is the number of nodes in the tree. The complexity comes from the sorting operation.
* Space Complexity:O(N), the size of`vals`.

```java
class Solution {
    List<Integer> vals;
    public int minDiffInBST(TreeNode root) {
        vals = new ArrayList();
        dfs(root);

        int ans = Integer.MAX_VALUE;
        for (int i = 0; i < vals.size() - 1; ++i)
            ans = Math.min(ans, vals.get(i+1) - vals.get(i));

        return ans;
    }

    public void dfs(TreeNode node) {
        if (node == null) return;
        dfs(node.left);
        vals.add(node.val);
        dfs(node.right);
    }
}
```

### Approach #2: In-Order Traversal <a href="#approach-2-inorder-traversal-accepted" id="approach-2-inorder-traversal-accepted"></a>

> In a binary search tree, an in-order traversal outputs the values of the tree in order. By remembering the previous value in this order, we could iterate over each possible difference, keeping the smallest one.

**Complexity Analysis**

* Time Complexity: O(N), whereNNis the number of nodes in the tree. We iterate over every node once.
* Space Complexity: O(H), whereHHis the height of the tree. This is the space used by the implicit call stack when calling`dfs`.

```java
class Solution {
    Integer prev, ans;
    public int minDiffInBST(TreeNode root) {
        prev = null;
        ans = Integer.MAX_VALUE;
        dfs(root);
        return ans;
    }

    public void dfs(TreeNode node) {
        if (node == null) return;
        dfs(node.left);
        if (prev != null)
            ans = Math.min(ans, node.val - prev);
        prev = node.val;
        dfs(node.right);
    }
}
```

Minimum Distance Between BST Nodes
