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.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDiffInBST(TreeNode* root) {
        const int inf = 1 << 30;
        int ans = inf, pre = -inf;
        auto dfs = [&](this auto&& dfs, TreeNode* root) -> void {
            if (!root) {
                return;
            }
            dfs(root->left);
            ans = min(ans, root->val - pre);
            pre = root->val;
            dfs(root->right);
        };
        dfs(root);
        return ans;
    }
};
