{ "data": { "question": { "questionId": "173", "questionFrontendId": "173", "categoryTitle": "Algorithms", "boundTopicId": 1548, "title": "Binary Search Tree Iterator", "titleSlug": "binary-search-tree-iterator", "content": "
Implement the BSTIterator
class that represents an iterator over the in-order traversal of a binary search tree (BST):
BSTIterator(TreeNode root)
Initializes an object of the BSTIterator
class. The root
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.boolean hasNext()
Returns true
if there exists a number in the traversal to the right of the pointer, otherwise returns false
.int next()
Moves the pointer to the right, then returns the number at the pointer.Notice that by initializing the pointer to a non-existent smallest number, the first call to next()
will return the smallest element in the BST.
You may assume that next()
calls will always be valid. That is, there will be at least a next number in the in-order traversal when next()
is called.
\n
Example 1:
\n\n\nInput\n["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]\n[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]\nOutput\n[null, 3, 7, true, 9, true, 15, true, 20, false]\n\nExplanation\nBSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);\nbSTIterator.next(); // return 3\nbSTIterator.next(); // return 7\nbSTIterator.hasNext(); // return True\nbSTIterator.next(); // return 9\nbSTIterator.hasNext(); // return True\nbSTIterator.next(); // return 15\nbSTIterator.hasNext(); // return True\nbSTIterator.next(); // return 20\nbSTIterator.hasNext(); // return False\n\n\n
\n
Constraints:
\n\n[1, 105]
.0 <= Node.val <= 106
105
calls will be made to hasNext
, and next
.\n
Follow up:
\n\nnext()
and hasNext()
to run in average O(1)
time and use O(h)
memory, where h
is the height of the tree?BSTIterator
,表示一个按中序遍历二叉搜索树(BST)的迭代器:\nBSTIterator(TreeNode root)
初始化 BSTIterator
类的一个对象。BST 的根节点 root
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。boolean hasNext()
如果向指针右侧遍历存在数字,则返回 true
;否则返回 false
。int next()
将指针向右移动,然后返回指针处的数字。注意,指针初始化为一个不存在于 BST 中的数字,所以对 next()
的首次调用将返回 BST 中的最小元素。
你可以假设 next()
调用总是有效的,也就是说,当调用 next()
时,BST 的中序遍历中至少存在一个下一个数字。
\n\n
示例:
\n\n\n输入\n[\"BSTIterator\", \"next\", \"next\", \"hasNext\", \"next\", \"hasNext\", \"next\", \"hasNext\", \"next\", \"hasNext\"]\n[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]\n输出\n[null, 3, 7, true, 9, true, 15, true, 20, false]\n\n解释\nBSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);\nbSTIterator.next(); // 返回 3\nbSTIterator.next(); // 返回 7\nbSTIterator.hasNext(); // 返回 True\nbSTIterator.next(); // 返回 9\nbSTIterator.hasNext(); // 返回 True\nbSTIterator.next(); // 返回 15\nbSTIterator.hasNext(); // 返回 True\nbSTIterator.next(); // 返回 20\nbSTIterator.hasNext(); // 返回 False\n\n\n
\n\n
提示:
\n\n[1, 105]
内0 <= Node.val <= 106
105
次 hasNext
和 next
操作\n\n
进阶:
\n\nnext()
和 hasNext()
操作均摊时间复杂度为 O(1)
,并使用 O(h)
内存。其中 h
是树的高度。\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n