Binary Search Tree Iterator (LC173)

06 May 2019

Problem description

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.

Solution

We can solve the problem by iterative binary tree inorder traversal

Below is python code

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

class BSTIterator:

    def __init__(self, root: TreeNode):
        self.stack = []
        current = root
        while current:
            self.stack.append(current)
            current = current.left

    def next(self) -> int:
        """
        @return the next smallest number
        """
        node = self.stack.pop()
        current = node.right
        while current:
            self.stack.append(current)
            current = current.left
        return node.val

    def hasNext(self) -> bool:
        """
        @return whether we have a next smallest number
        """
        return len(self.stack) != 0


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