Minimum Distance Between BST Nodes
<p>Given the <code>root</code> of a Binary Search Tree (BST), return <em>the minimum difference between the values of any two different nodes in the tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0783.Minimum%20Distance%20Between%20BST%20Nodes/images/bst1.jpg" style="width: 292px; height: 301px;" />
<pre>
<strong>Input:</strong> root = [4,2,6,1,3]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0783.Minimum%20Distance%20Between%20BST%20Nodes/images/bst2.jpg" style="width: 282px; height: 301px;" />
<pre>
<strong>Input:</strong> root = [1,0,48,null,null,12,49]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 100]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Note:</strong> This question is the same as 530: <a href="https://leetcode.com/problems/minimum-absolute-difference-in-bst/" target="_blank">https://leetcode.com/problems/minimum-absolute-difference-in-bst/</a></p>
Tree; Depth-First Search; Breadth-First Search; Binary Search Tree; Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function minDiffInBST(root: TreeNode | null): number {
    let [ans, pre] = [Infinity, -Infinity];
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, root.val - pre);
        pre = root.val;
        dfs(root.right);
    };
    dfs(root);
    return ans;
}
