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173.binary-search-tree-iterator.py
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79 lines (72 loc) · 2.8 KB
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# Tag: Stack, Tree, Design, Binary Search Tree, Binary Tree, Iterator
# Time: O(1)
# Space: O(H)
# Ref: -
# Note: InOrder
# 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.
#
# Example 1:
#
#
# Input
# ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
# [[[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
#
#
# Constraints:
#
# The number of nodes in the tree is in the range [1, 105].
# 0 <= Node.val <= 106
# At most 105 calls will be made to hasNext, and next.
#
#
# Follow up:
#
# 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?
#
#
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class BSTIterator:
def __init__(self, root: Optional[TreeNode]):
self.cur = root
self.stack = []
def next(self) -> int:
while (self.cur):
self.stack.append(self.cur)
self.cur = self.cur.left
self.cur = self.stack.pop()
res = self.cur.val
self.cur = self.cur.right
return res
def hasNext(self) -> bool:
return (self.cur is not None or len(self.stack) > 0)
# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()