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LeetCode: Binary Search Tree Iterator

Posted on January 13, 2018July 26, 2020 by braindenny

Binary Search Tree Iterator



Similar Problems:

  • CheatSheet: Leetcode For Code Interview
  • CheatSheet: Common Code Problems & Follow-ups
  • Tag: #oodesign

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## https://code.dennyzhang.com/binary-search-tree-iterator
## Basic Ideas: In-order traversal. Use a stack
##      Store directed left children from root.
##      When next() be called, pop one element and process its right child as new root.
## Complexity:
##
# Definition for a  binary tree node
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class BSTIterator(object):
    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.stack = []
        p = root
        while p:
            self.stack.append(p)
            p = p.left

    def hasNext(self):
        """
        :rtype: bool
        """
        return len(self.stack) > 0

    def next(self):
        """
        :rtype: int
        """
        node = self.stack.pop()
        p = node.right
        while p:
            self.stack.append(p)
            p = p.left
        return node.val

# Your BSTIterator will be called like this:
# i, v = BSTIterator(root), []
# while i.hasNext(): v.append(i.next())
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Posted in MediumTagged #iterator, oodesign

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