Review the fundamentals: Binary Trees guide and time complexities.
ask us to return — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Trees and solve the ask us to return problem optimally.
O(N)O(W)Problem Description
You are given a binary tree represented by its root node. The tree is not necessarily balanced, and each node contains an integer value. Your task is to determine the index of the first node (in level-order traversal) whose value matches a specific target integer. If the target value does not exist in the tree, return -1.
Level-order traversal visits nodes from top to bottom, left to right. The index starts at 0 for the root node. For a node at index i, its left child is at index 2*i + 1 and its right child is at index 2*i + 2, assuming a complete binary tree structure. However, since the tree may be incomplete, you must simulate the traversal using a queue to correctly assign indices based on the actual structure.
Return the 0-based index of the first occurrence of the target value in the level-order traversal. If the target is not found, return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"ask us to return"
WHY DOES IT MATTER?
Breadth‑first search is the canonical pattern for problems that require the shortest‑path or earliest‑encounter guarantee in hierarchical structures. It ensures the first occurrence of a condition is found with minimal work, which is critical for real‑time systems and large‑scale data processing.
OPTIMIZATION CHALLENGE
The key insight is to avoid traversing the entire tree when a match is found early. By using a queue, we can stop as soon as the first matching node is dequeued, achieving O(N) worst‑case but often much less in practice, while keeping auxiliary memory bounded by the tree's maximum width.
REAL-WORLD CONNECTION
Think of a customer support ticket system where tickets are organized by priority levels; BFS corresponds to checking tickets level by level, guaranteeing that the highest‑priority (shallowest) matching ticket is addressed first, analogous to routing packets in a network layer by layer.
During an interview, implement the BFS loop first, then add the index counter and early‑exit condition. Keep the code modular: a helper to enqueue children only if they exist, and always check for null root up front to avoid edge‑case bugs.
COMPLEXITY AT A GLANCE
O(N)O(W)Core Theory — Why This Approach?
Level‑order traversal (Breadth‑First Search) visits a binary tree node‑by‑node from the root, proceeding level by level and left to right within each level. By using a FIFO queue we can guarantee that nodes are processed in exactly the order required to assign a deterministic index to each visited node. The index of the first node whose value equals the target is simply the position of that node in the BFS sequence, starting at 0 (or 1, depending on convention).\n\nA naïve approach might attempt a depth‑first recursion (pre‑order, in‑order, or post‑order) and keep a global counter, but such traversals do not respect the level‑order guarantee; they could encounter a matching node deeper in the tree before a shallower match, leading to an incorrect index. Moreover, recursive DFS can overflow the call stack on highly unbalanced trees. The optimal paradigm is an iterative BFS using a queue, which processes each node exactly once, maintains the correct ordering, and uses space proportional only to the maximum width of the tree rather than its depth.
Interview Questions on This Problem
Q1How would you adapt the solution to return the indices of *all* nodes that match the target value instead of just the first one?
Continue the BFS until the queue is empty, appending the current index to a result list whenever node.val == target. The algorithm remains O(N) time and O(W) space, with the additional O(K) space for K matches.
Q2If the tree is stored in a distributed key‑value store where child pointers are remote calls, how can you minimize network latency while searching for the target?
Batch fetch all nodes at the current level in a single request before moving to the next level, effectively performing a level‑wise bulk read. This preserves BFS order while reducing round‑trip overhead to O(depth) network calls.
Q3What changes are needed if the index should be 1‑based instead of 0‑based, and how would you handle an empty tree?
Initialize the counter at 1 and increment after each node is processed; if root is null, immediately return -1 because there are no nodes to index.
Examples
Input
root = [1, 2, 3, 4, 5, 6, 7], target = 5
Output
4
Explanation: Level-order traversal: [1 (index 0), 2 (index 1), 3 (index 2), 4 (index 3), 5 (index 4), 6 (index 5), 7 (index 6)]. The target 5 is found at index 4.
Input
root = [10, 20, 30, null, 40, 50, 60], target = 40
Output
3
Explanation: Level-order traversal: [10 (index 0), 20 (index 1), 30 (index 2), 40 (index 3), 50 (index 4), 60 (index 5)]. The target 40 is found at index 3.
Input
root = [1, 2, 3, 4, 5, 6, 7], target = 8
Output
-1
Explanation: Level-order traversal: [1 (index 0), 2 (index 1), 3 (index 2), 4 (index 3), 5 (index 4), 6 (index 5), 7 (index 6)]. The target 8 is not found in the tree, so return -1.
Input
root = [1, null, 2, null, 3, null, 4], target = 3
Output
4
Explanation: Level-order traversal: [1 (index 0), 2 (index 2), 3 (index 4), 4 (index 6)]. The target 3 is found at index 4.
Constraints
- 1 <= number of nodes in the tree <= 10^5
- -10^9 <= node.val <= 10^9
- The tree is a valid binary tree (each node has at most two children).
- The target value is an integer within the range of node values.
Optimal Approach & Strategy
Use an iterative BFS with a queue, incrementing an index as each node is dequeued, and return immediately when the node's value equals the target.
Brute Force Approach
Perform a depth‑first traversal (pre‑order) while maintaining a global counter; after the traversal finishes, scan the recorded order to find the first matching index.
Code Solutions
function findConsecutiveStart(arr, k) {
if (!arr || arr.length < k || k <= 0) return -1;
let count = 1;
for (let i = 1; i < arr.length; i++) {
if (arr[i] === arr[i - 1]) {
count++;
} else {
count = 1;
}
if (count === k) {
return i - k + 1;
}
}
return k === 1 ? 0 : -1;
}#include <vector>
using namespace std;
class Solution {
public:
int findConsecutiveStart(vector<int>& arr, int k) {
if (arr.empty() || k <= 0 || arr.size() < static_cast<size_t>(k)) return -1;
if (k == 1) return 0;
int count = 1;
for (size_t i = 1; i < arr.size(); ++i) {
if (arr[i] == arr[i - 1]) {
count++;
} else {
count = 1;
}
if (count == k) {
return i - k + 1;
}
}
return -1;
}
};public class Solution {
public int findConsecutiveStart(int[] arr, int k) {
if (arr == null || arr.length < k || k <= 0) return -1;
if (k == 1) return 0;
int count = 1;
for (int i = 1; i < arr.length; i++) {
if (arr[i] == arr[i - 1]) {
count++;
if (count == k) return i - k + 1;
} else {
count = 1;
}
}
return -1;
}
}def find_consecutive_start(arr: list[int], k: int) -> int:
if not arr or k <= 0 or len(arr) < k:
return -1
if k == 1:
return 0
count = 1
for i in range(1, len(arr)):
if arr[i] == arr[i - 1]:
count += 1
else:
count = 1
if count == k:
return i - k + 1
return -1function findConsecutiveStart(arr, k) {
if (!arr || arr.length < k || k <= 0) return -1;
let count = 1;
for (let i = 1; i < arr.length; i++) {
if (arr[i] === arr[i - 1]) {
count++;
} else {
count = 1;
}
if (count === k) {
return i - k + 1;
}
}
return k === 1 ? 0 : -1;
}Solve in Interative Editor
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