Binary Search Tree Iterator(tree inorder)

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Callingnext()will return the next smallest number in the BST.

Note:next()andhasNext()should run in average O(1) time and uses O(h) memory, wherehis the height of the tree.

分析

inorder

iterative: hasnext里是while的条件 next里是遍历的过程。要拆开。想象不断调用hasnext 就是不断while

recursive的话就是直接Inorder做完存在List里。然后控制List的index

follow up: post order，一样遍历，就是捕捉stack pop的瞬间

# 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.cur = root

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.cur or self.stack


    def next(self):
        """
        :rtype: int
        """

        while self.cur:
            self.stack.append(self.cur)
            self.cur = self.cur.left
        res = self.stack.pop()
        self.cur = res.right

        return res.val




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