leetcode/binary_search_tree_iterator.py at main · atib92/leetcode · GitHub

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'''
Binary Search Tree Iterator
Solved
Medium
Topics
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Companies
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?
'''

'''
Algo: Keep a stack of nodes. When we pop a node, we push all the left children of the right child of the popped node.
This serves the purpose of in-order traversal. The stack will always have the next node to be returned at the top.

Applications: two sum on bst w/o having to convert it to an array, find the kth smallest element in a bst, find the closest value to a target in a bst, etc.
'''


# 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._stack = []
        self._tos = -1
        self._push_left(root)

    def _push_left(self, node):
        while node:
            self._tos += 1
            self._stack.append(node)
            node = node.left


    def next(self) -> int:
        node = self._stack.pop()
        self._tos -= 1
        self._push_left(node.right)
        return node.val

    def hasNext(self) -> bool:
        return self._tos > -1


# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()