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Description
tab: English
<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> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png" style="width: 189px; height: 178px;" />
<pre>
<strong>Input</strong>
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[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> </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 <= Node.val <= 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> </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 <code>O(h)</code> memory, where <code>h</code> is the height of the tree?</li>
</ul>
---
[submissions](https://leetcode.com/problems/binary-search-tree-iterator/submissions/) | [solutions](https://leetcode.com/problems/binary-search-tree-iterator/solutions/)
tab: 中文
实现一个二叉搜索树迭代器类<code>BSTIterator</code> ，表示一个按中序遍历二叉搜索树（BST）的迭代器：
<div class="original__bRMd">
<div>
<ul>
<li><code>BSTIterator(TreeNode root)</code> 初始化 <code>BSTIterator</code> 类的一个对象。BST 的根节点 <code>root</code> 会作为构造函数的一部分给出。指针应初始化为一个不存在于 BST 中的数字，且该数字小于 BST 中的任何元素。</li>
<li><code>boolean hasNext()</code> 如果向指针右侧遍历存在数字，则返回 <code>true</code> ；否则返回 <code>false</code> 。</li>
<li><code>int next()</code>将指针向右移动，然后返回指针处的数字。</li>
</ul>
<p>注意，指针初始化为一个不存在于 BST 中的数字，所以对 <code>next()</code> 的首次调用将返回 BST 中的最小元素。</p>
</div>
</div>
<p>你可以假设 <code>next()</code> 调用总是有效的，也就是说，当调用 <code>next()</code> 时，BST 的中序遍历中至少存在一个下一个数字。</p>
<p> </p>
<p><strong>示例：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png" style="width: 189px; height: 178px;" />
<pre>
<strong>输入</strong>
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
<strong>输出</strong>
[null, 3, 7, true, 9, true, 15, true, 20, false]
<strong>解释</strong>
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // 返回 3
bSTIterator.next(); // 返回 7
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 9
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 15
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 20
bSTIterator.hasNext(); // 返回 False
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点的数目在范围 <code>[1, 10<sup>5</sup>]</code> 内</li>
<li><code>0 <= Node.val <= 10<sup>6</sup></code></li>
<li>最多调用 <code>10<sup>5</sup></code> 次 <code>hasNext</code> 和 <code>next</code> 操作</li>
</ul>
<p> </p>
<p><strong>进阶：</strong></p>
<ul>
<li>你可以设计一个满足下述条件的解决方案吗？<code>next()</code> 和 <code>hasNext()</code> 操作均摊时间复杂度为 <code>O(1)</code> ，并使用 <code>O(h)</code> 内存。其中 <code>h</code> 是树的高度。</li>
</ul>
---
[提交记录](https://leetcode.cn/problems/binary-search-tree-iterator/submissions/) | [题解](https://leetcode.cn/problems/binary-search-tree-iterator/solution/)
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