173. Binary Search Tree Iterator

题目

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

这是一道设计题，相对来说很不熟悉，但归根结底是BST的问题。要实现这个迭代器，我们用到一个stack，先把root左边的所有节点放进栈，然后依次把每个访问到的节点的右子树的左边放进栈。

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())