Solution.py

"""
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.

 

Example:
![img](https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png)


BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false
 

Note:

next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.
"""


# 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.root = root
        stack = []
        curr = root
        while curr is not None:
            stack.append(curr)
            curr = curr.left
        self.stack = stack
        

    def next(self):
        """
        @return the next smallest number
        :rtype: int
        """
        curr = self.stack.pop()
        t = curr.right
        while t is not None:
            self.stack.append(t)
            t = t.left
        return curr.val
        

    def hasNext(self):
        """
        @return whether we have a next smallest number
        :rtype: bool
        """
        return 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()