LeetCode #1261 — MEDIUM

Find Elements in a Contaminated Binary Tree

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode

The Problem

Problem Statement

Given a binary tree with the following rules:

root.val == 0
For any treeNode:
1. If treeNode.val has a value x and treeNode.left != null, then treeNode.left.val == 2 * x + 1
2. 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.
bool find(int target) Returns true if the target value exists in the recovered binary tree.

Example 1:

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

Example 2:

Input
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
Output
[null,true,true,false]
Explanation
FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
findElements.find(1); // return True
findElements.find(3); // return True
findElements.find(5); // return False

Example 3:

Input
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
Output
[null,true,false,false,true]
Explanation
FindElements findElements = new FindElements([-1,null,-1,-1,null,-1]);
findElements.find(2); // return True
findElements.find(3); // return False
findElements.find(4); // return False
findElements.find(5); // return True

Constraints:

TreeNode.val == -1
The height of the binary tree is less than or equal to 20
The total number of nodes is between [1, 10⁴]
Total calls of find() is between [1, 10⁴]
0 <= target <= 10⁶

Step 01

Brute Force Baseline

Problem summary: 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. bool find(int target) Returns true if the target value exists in the recovered binary tree.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Tree · Design

Example 1

["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]

Example 2

["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]

Example 3

["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]

Step 02

Core Insight

What unlocks the optimal approach

Use DFS to traverse the binary tree and recover it.
Use a hashset to store TreeNode.val for finding.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1261: Find Elements in a Contaminated Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class FindElements {
    private Set<Integer> s = new HashSet<>();

    public FindElements(TreeNode root) {
        root.val = 0;
        dfs(root);
    }

    public boolean find(int target) {
        return s.contains(target);
    }

    private void dfs(TreeNode root) {
        s.add(root.val);
        if (root.left != null) {
            root.left.val = root.val * 2 + 1;
            dfs(root.left);
        }
        if (root.right != null) {
            root.right.val = root.val * 2 + 2;
            dfs(root.right);
        }
    }
}

/**
 * Your FindElements object will be instantiated and called as such:
 * FindElements obj = new FindElements(root);
 * boolean param_1 = obj.find(target);
 */

// Accepted solution for LeetCode #1261: Find Elements in a Contaminated Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
type FindElements struct {
	s map[int]bool
}

func Constructor(root *TreeNode) FindElements {
	root.Val = 0
	s := map[int]bool{}
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		s[root.Val] = true
		if root.Left != nil {
			root.Left.Val = root.Val*2 + 1
			dfs(root.Left)
		}
		if root.Right != nil {
			root.Right.Val = root.Val*2 + 2
			dfs(root.Right)
		}
	}
	dfs(root)
	return FindElements{s}
}

func (this *FindElements) Find(target int) bool {
	return this.s[target]
}

/**
 * Your FindElements object will be instantiated and called as such:
 * obj := Constructor(root);
 * param_1 := obj.Find(target);
 */

# Accepted solution for LeetCode #1261: Find Elements in a Contaminated Binary Tree
# 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)

// Accepted solution for LeetCode #1261: Find Elements in a Contaminated Binary Tree
use rustgym_util::*;
use std::collections::HashSet;

trait Preorder {
    fn recover(&mut self, x: i32, hs: &mut HashSet<i32>);
}

impl Preorder for TreeLink {
    fn recover(&mut self, x: i32, hs: &mut HashSet<i32>) {
        if let Some(node) = self {
            hs.insert(x);
            node.borrow_mut().val = x;
            node.borrow_mut().left.recover(2 * x + 1, hs);
            node.borrow_mut().right.recover(2 * x + 2, hs);
        }
    }
}

struct FindElements {
    root: TreeLink,
    hs: HashSet<i32>,
}

impl FindElements {
    fn new(mut root: TreeLink) -> Self {
        let mut hs = HashSet::new();
        root.recover(0, &mut hs);
        FindElements { root, hs }
    }

    fn find(&self, target: i32) -> bool {
        self.hs.contains(&target)
    }
}

#[test]
fn test() {
    let root = tree!(-1, None, tree!(-1));
    let fe = FindElements::new(root);
    let target = 1;
    let res = false;
    assert_eq!(fe.find(target), res);
    let target = 2;
    let res = true;
    assert_eq!(fe.find(target), res);
    let root = tree!(-1, tree!(-1, tree!(-1), tree!(-1)), tree!(-1));
    let fe = FindElements::new(root);
    let target = 1;
    let res = true;
    assert_eq!(fe.find(target), res);
    let target = 3;
    let res = true;
    assert_eq!(fe.find(target), res);
    let target = 5;
    let res = false;
    assert_eq!(fe.find(target), res);
    let root = tree!(-1, None, tree!(-1, tree!(-1, tree!(-1), None), None));
    let fe = FindElements::new(root);
    let target = 2;
    let res = true;
    assert_eq!(fe.find(target), res);
    let target = 3;
    let res = false;
    assert_eq!(fe.find(target), res);
    let target = 4;
    let res = false;
    assert_eq!(fe.find(target), res);
    let target = 5;
    let res = true;
    assert_eq!(fe.find(target), res);
}

// Accepted solution for LeetCode #1261: Find Elements in a Contaminated Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

class FindElements {
    readonly #s = new Set<number>();

    constructor(root: TreeNode | null) {
        root.val = 0;

        const dfs = (node: TreeNode | null, x = 0) => {
            if (!node) return;

            this.#s.add(x);
            dfs(node.left, x * 2 + 1);
            dfs(node.right, x * 2 + 2);
        };

        dfs(root);
    }

    find(target: number): boolean {
        return this.#s.has(target);
    }
}

/**
 * Your FindElements object will be instantiated and called as such:
 * var obj = new FindElements(root)
 * var param_1 = obj.find(target)
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(h)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.