Leetcode #173: Binary Search Tree Iterator
In this guide, we solve Leetcode #173 Binary Search Tree Iterator in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST): BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Tree, Design, Binary Search Tree, Binary Tree, Iterator
Intuition
The problem has a natural nested or last-in-first-out structure.
A stack lets us resolve matches in the correct order as we scan.
Approach
Push items as they appear and pop when you can finalize a decision.
The stack captures the unresolved part of the input.
Steps:
- Push elements as you scan.
- Pop when a rule or match is satisfied.
- Use the stack to compute results.
Example
Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]
Explanation
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // return 3
bSTIterator.next(); // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 20
bSTIterator.hasNext(); // return False
Python 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):
def inorder(root):
if root:
inorder(root.left)
self.vals.append(root.val)
inorder(root.right)
self.cur = 0
self.vals = []
inorder(root)
def next(self) -> int:
res = self.vals[self.cur]
self.cur += 1
return res
def hasNext(self) -> bool:
return self.cur < len(self.vals)
# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.