Inorder Successor in BST
<p>Given the <code>root</code> of a binary search tree and a node <code>p</code> in it, return <em>the in-order successor of that node in the BST</em>. If the given node has no in-order successor in the tree, return <code>null</code>.</p>

<p>The successor of a node <code>p</code> is the node with the smallest key greater than <code>p.val</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_1.png" style="width: 122px; height: 117px;" />
<pre>
<strong>Input:</strong> root = [2,1,3], p = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> 1&#39;s in-order successor node is 2. Note that both p and the return value is of TreeNode type.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_2.png" style="width: 246px; height: 229px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,null,1], p = 6
<strong>Output:</strong> null
<strong>Explanation:</strong> There is no in-order successor of the current node, so the answer is <code>null</code>.
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
	<li>All Nodes will have unique values.</li>
</ul>
Tree; Depth-First Search; Binary Search Tree; Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func inorderSuccessor(root *TreeNode, p *TreeNode) (ans *TreeNode) {
	for root != nil {
		if root.Val > p.Val {
			ans = root
			root = root.Left
		} else {
			root = root.Right
		}
	}
	return
}