https://leetcode.com/problems/binary-search-tree-iterator/#/description

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

Sol：

https://discuss.leetcode.com/topic/6575/my-solutions-in-3-languages-with-stack

I use Stack to store directed left children from root.
When next() be called, I just pop one element and process its right child as new root.
The code is pretty straightforward.

https://discuss.leetcode.com/topic/6629/two-python-solutions-stack-and-generator

# 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
        """
        # stack stores all left nodes
        self.stack = []
        while root:
            self.stack.append(root)
            root = root.left
        
        
        
    # @return a boolean, whether we have a next smallest number
    def hasNext(self):
        """
        :rtype: bool
        """
        
        return len(self.stack) > 0
        
    
    # @return an integer, the next smallest number
    def next(self):
        """
        :rtype: int
        """
        
        # node is the left leaf from bottom 
        node = self.stack.pop()
        x = node.right
        while x:
            self.stack.append(x)
            x = x.left
        return node.val
        
        
        

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

173. Binary Search Tree Iterator - Unsolved