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

實現思路

要找到二叉搜尋樹中的最小節點，應該從根節點遞迴遍歷左節點，直到遍歷的節點沒有左節點，那麼該節點就是二叉樹中的最小節點。現在已經有二叉搜尋樹中沒有訪問過的最小節點了，那麼當訪問了該節點後，剩餘沒有訪問的樹中最小的節點在哪裡呢？如果該節點有右子樹，那麼在它的右子樹中（又回到了找一棵二叉搜尋樹的最小元素，不過這棵二叉搜尋樹變小了）；如果沒有右子樹，那麼就是它的父節點。

為了能夠快速定位到父節點，我們可以用棧將遍歷路徑暫存起來，當進行next()操作時，我們彈出棧頂元素並進行訪問，如果它有右子樹的話就遍歷它的右子樹；如果沒有右子樹，當下次出棧操作時就是訪問當前節點的父節點了。

實現程式碼

# 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._pushLeft(root)
        

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

    def next(self):
        """
        :rtype: int
        """
        node = self.stack.pop()
        self._pushLeft(node.right)
        return node.val
    
    
    def _pushLeft(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())