Leetcode 173. Binary Search Tree Iterator

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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.

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

Answer:

# 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 dfs(self,node):
        if node==None:
            return
        left=node.left
        right=node.right
        self.dfs(left)
        self.nodelist.append(node)
        self.dfs(right)
        
    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.nodelist=[]
        self.dfs(root)
        self.pos=0

    def hasNext(self):
        """
        :rtype: bool
        """
        if self.pos<len(self.nodelist):
            return True
        return False

    def next(self):
        """
        :rtype: int
        """
        result=self.nodelist[self.pos].val
        self.pos+=1
        return result

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

inoder traversal