Leetcode #1261: Find Elements in a Contaminated Binary Tree
In this guide, we solve Leetcode #1261 Find Elements in a Contaminated Binary Tree 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 a binary tree with the following rules: root.val == 0 For any treeNode: If treeNode.val has a value x and treeNode.left != null, then treeNode.left.val == 2 * x + 1 If treeNode.val has a value x and treeNode.right != null, then treeNode.right.val == 2 * x + 2 Now the binary tree is contaminated, which means all treeNode.val have been changed to -1. Implement the FindElements class: FindElements(TreeNode* root) Initializes the object with a contaminated binary tree and recovers it.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Design, Hash Table, 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
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
Output
[null,false,true]
Explanation
FindElements findElements = new FindElements([-1,null,-1]);
findElements.find(1); // return False
findElements.find(2); // return 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 FindElements:
def __init__(self, root: Optional[TreeNode]):
def dfs(root: Optional[TreeNode]):
self.s.add(root.val)
if root.left:
root.left.val = root.val * 2 + 1
dfs(root.left)
if root.right:
root.right.val = root.val * 2 + 2
dfs(root.right)
root.val = 0
self.s = set()
dfs(root)
def find(self, target: int) -> bool:
return target in self.s
# Your FindElements object will be instantiated and called as such:
# obj = FindElements(root)
# param_1 = obj.find(target)
Complexity
The time complexity is O(n). The space complexity is , where is the number of nodes in the binary tree.
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.