題目
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.
思路
這是一道設計題荆陆，相對來說很不熟悉，但歸根結底是BST的問題巷怜。要實現這個迭代器，我們用到一個stack水由，先把root左邊的所有節(jié)點放進棧贪磺，然后依次把每個訪問到的節(jié)點的右子樹的左邊放進棧楞艾。
Python
# 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 = []
self.pushAll(root)
def hasNext(self):
"""
:rtype: bool
"""
return self.stack
def next(self):
"""
:rtype: int
"""
temp = self.stack.pop()
self.pushAll(temp.right)
return temp.val
def pushAll(self, node):
while node:
self.stack.append(node)
node = node.left
# Your BSTIterator will be called like this:
# i, v = BSTIterator(root), []
# while i.hasNext(): v.append(i.next())
