## 173.Binary Search Tree Iterator

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, wherehis the height of the tree.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

Thoughts:

1. Using stack to record path.
2. When initialized, traverse to the leftmost
3. when retrieving, pop out from the stack and use its right child (if it does have) as a root node to put more nodes in front of the stack (In order traversal)

Code Python T:O(1) S:O(h)

``````# 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 = []
cur = root
while cur is not None:
self.stack.append(cur)
if cur.left:
cur = cur.left
else:
break
# print("len(self.stack) %s" %(len(self.stack)))
def hasNext(self):
"""
:rtype: bool
"""
return self.stack

def next(self):
"""
:rtype: int
"""
if not self.stack: return None

ans = self.stack.pop()
cur = ans
# check whether there is a right child, if there is, traverse towards the leftmost
# of the right child.
if cur.right:
cur = cur.right
while cur:
self.stack.append(cur)
if cur.left:
cur = cur.left
else:
break

return ans.val

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

from siyang2's post