README.md

173. Binary Search Tree Iterator

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.

Example:

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false

Note:

• next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
• You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

Solutions (Python)

1. Inorder Traversal

# 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 = []
        self.curr = root

        while self.curr:
            self.stack.append(self.curr)
            self.curr = self.curr.left

    def next(self) -> int:
        """
        @return the next smallest number
        """
        self.curr = self.stack.pop()
        ret = self.curr.val

        self.curr = self.curr.right
        while self.curr:
            self.stack.append(self.curr)
            self.curr = self.curr.left

        return ret

    def hasNext(self) -> bool:
        """
        @return whether we have a next smallest number
        """
        return self.stack


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