Leetcode #653: Two Sum IV - Input is a BST
In this guide, we solve Leetcode #653 Two Sum IV - Input is a BST 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
Given the root of a binary search tree and an integer k, return true if there exist two elements in the BST such that their sum is equal to k, or false otherwise. Example 1: Input: root = [5,3,6,2,4,null,7], k = 9 Output: true Example 2: Input: root = [5,3,6,2,4,null,7], k = 28 Output: false Constraints: The number of nodes in the tree is in the range [1, 104].
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Binary Search Tree, Hash Table, Two Pointers, Binary Tree
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: root = [5,3,6,2,4,null,7], k = 9
Output: true
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 Solution:
def findTarget(self, root: Optional[TreeNode], k: int) -> bool:
def dfs(root):
if root is None:
return False
if k - root.val in vis:
return True
vis.add(root.val)
return dfs(root.left) or dfs(root.right)
vis = set()
return dfs(root)
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.