[LeetCode] 173. Binary Search Tree Iterator

Problem : https://leetcode.com/problems/binary-search-tree-iterator/

Because in-order traverse BST can generate a ascending sequence.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class BSTIterator:

    def __init__(self, root: TreeNode):
        
        def inorder(node):
            if node:
                yield from inorder(node.right)
                yield node.val
                yield from inorder(node.left)
        
        self.nodes = list(inorder(root))

    def next(self) -> int:
        return self.nodes.pop()

    def hasNext(self) -> bool:
        return self.nodes


# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()

A stack based iterative solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class BSTIterator:

    def __init__(self, root: TreeNode):
        self.stack = [root] if root else []
        

    def next(self) -> int:
        node = self.stack[-1]
        
        while node.left:
            self.stack.append(node.left)
            tmp = node.left
            node.left = None
            node = tmp
        
        # pop the parent node
        self.stack.pop()
        
        if node.right:
            self.stack.append(node.right)
            node.right = None
        
        return node.val
        

    def hasNext(self) -> bool:
        return self.stack
        


# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()

Edited on 04/22/2021. Refactor the recursive and iterative solution.