binary-search-tree-iterator

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Algorithms

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

<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>

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.

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.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png" style="width: 189px; height: 178px;" />
<pre>
Input
[&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]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation
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>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 105]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 106</code></li>
	<li>At most <code>105</code> calls will be made to <code>hasNext</code>, and <code>next</code>.</li>
</ul>

&nbsp;
Follow up:

<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>

Binary Search Tree Iterator

Binary Tree Inorder Traversal

binary-tree-inorder-traversal

二叉树的中序遍历

Flatten 2D Vector

flatten-2d-vector

展开二维向量

Zigzag Iterator

zigzag-iterator

锯齿迭代器

Peeking Iterator

peeking-iterator

窥视迭代器

Inorder Successor in BST

inorder-successor-in-bst

二叉搜索树中的中序后继

Binary Search Tree Iterator II

binary-search-tree-iterator-ii

二叉搜索树迭代器 II

Stack

Tree

Design

Binary Search Tree

Binary Tree

Iterator

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>

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。
</div>
</div>

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。

 

示例：
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png" style="width: 189px; height: 178px;" />
<pre>
输入
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
输出
[null, 3, 7, true, 9, true, 15, true, 20, false]

解释
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>

 

提示：

<ul>
	<li>树中节点的数目在范围 <code>[1, 105]</code> 内</li>
	<li><code>0 <= Node.val <= 106</code></li>
	<li>最多调用 <code>105</code> 次 <code>hasNext</code> 和 <code>next</code> 操作</li>
</ul>

 

进阶：

<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>