Two Sum IV - Input is a BST
<p>Given the <code>root</code> of a binary search tree and an integer <code>k</code>, return <code>true</code> <em>if there exist two elements in the BST such that their sum is equal to</em> <code>k</code>, <em>or</em> <code>false</code> <em>otherwise</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/0600-0699/0653.Two%20Sum%20IV%20-%20Input%20is%20a%20BST/images/sum_tree_1.jpg" style="width: 400px; height: 229px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,7], k = 9
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0653.Two%20Sum%20IV%20-%20Input%20is%20a%20BST/images/sum_tree_2.jpg" style="width: 400px; height: 229px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,7], k = 28
<strong>Output:</strong> false
</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>4</sup> &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
	<li><code>root</code> is guaranteed to be a <strong>valid</strong> binary search tree.</li>
	<li><code>-10<sup>5</sup> &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>
Tree; Depth-First Search; Breadth-First Search; Binary Search Tree; Hash Table; Two Pointers; 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 findTarget(root: TreeNode | null, k: number): boolean {
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return false;
        }
        if (vis.has(k - root.val)) {
            return true;
        }
        vis.add(root.val);
        return dfs(root.left) || dfs(root.right);
    };
    const vis = new Set<number>();
    return dfs(root);
}
