Binary Search Tree Iterator
<p>Implement the <code>BSTIterator</code> class that represents an iterator over the <strong><a href="https://en.wikipedia.org/wiki/Tree_traversal#In-order_(LNR)" target="_blank">in-order traversal</a></strong> of a binary search tree (BST):</p>

<ul>
	<li><code>BSTIterator(TreeNode root)</code> Initializes an object of the <code>BSTIterator</code> class. The <code>root</code> of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.</li>
	<li><code>boolean hasNext()</code> Returns <code>true</code> if there exists a number in the traversal to the right of the pointer, otherwise returns <code>false</code>.</li>
	<li><code>int next()</code> Moves the pointer to the right, then returns the number at the pointer.</li>
</ul>

<p>Notice that by initializing the pointer to a non-existent smallest number, the first call to <code>next()</code> will return the smallest element in the BST.</p>

<p>You may assume that <code>next()</code> calls will always be valid. That is, there will be at least a next number in the in-order traversal when <code>next()</code> is called.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0173.Binary%20Search%20Tree%20Iterator/images/bst-tree.png" style="width: 189px; height: 178px;" />
<pre>
<strong>Input</strong>
[&quot;BSTIterator&quot;, &quot;next&quot;, &quot;next&quot;, &quot;hasNext&quot;, &quot;next&quot;, &quot;hasNext&quot;, &quot;next&quot;, &quot;hasNext&quot;, &quot;next&quot;, &quot;hasNext&quot;]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
<strong>Output</strong>
[null, 3, 7, true, 9, true, 15, true, 20, false]

<strong>Explanation</strong>
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // return 3
bSTIterator.next(); // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 20
bSTIterator.hasNext(); // return 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>5</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>6</sup></code></li>
	<li>At most <code>10<sup>5</sup></code> calls will be made to <code>hasNext</code>, and <code>next</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<ul>
	<li>Could you implement <code>next()</code> and <code>hasNext()</code> to run in average <code>O(1)</code> time and use&nbsp;<code>O(h)</code> memory, where <code>h</code> is the height of the tree?</li>
</ul>
Stack; Tree; Design; Binary Search Tree; Binary Tree; Iterator
/**
 * 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 BSTIterator {
public:
    vector<int> vals;
    int cur;
    BSTIterator(TreeNode* root) {
        cur = 0;
        inorder(root);
    }

    int next() {
        return vals[cur++];
    }

    bool hasNext() {
        return cur < vals.size();
    }

    void inorder(TreeNode* root) {
        if (root) {
            inorder(root->left);
            vals.push_back(root->val);
            inorder(root->right);
        }
    }
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */