42 lines
2.0 KiB
Markdown
42 lines
2.0 KiB
Markdown
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Implement the `BSTIterator` class that represents an iterator over the **[in-order traversal](https://en.wikipedia.org/wiki/Tree_traversal#In-order_(LNR))** of a binary search tree (BST):
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* `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.
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* `boolean hasNext()` Returns `true` if there exists a number in the traversal to the right of the pointer, otherwise returns `false`.
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* `int next()` Moves the pointer to the right, then returns the number at the pointer.
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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.
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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.
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**Example 1:**
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Input
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["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
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[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
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Output
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[null, 3, 7, true, 9, true, 15, true, 20, false]
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Explanation
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BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
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bSTIterator.next(); // return 3
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bSTIterator.next(); // return 7
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bSTIterator.hasNext(); // return True
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bSTIterator.next(); // return 9
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bSTIterator.hasNext(); // return True
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bSTIterator.next(); // return 15
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bSTIterator.hasNext(); // return True
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bSTIterator.next(); // return 20
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bSTIterator.hasNext(); // return False
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**Constraints:**
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* The number of nodes in the tree is in the range `[1, 105]`.
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* `0 <= Node.val <= 106`
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* At most `105` calls will be made to `hasNext`, and `next`.
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**Follow up:**
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* Could you implement `next()` and `hasNext()` to run in average `O(1)` time and use `O(h)` memory, where `h` is the height of the tree?
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