LeetCode #2509 — HARD

Cycle Length Queries in a Tree

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an integer n. There is a complete binary tree with 2n - 1 nodes. The root of that tree is the node with the value 1, and every node with a value val in the range [1, 2n - 1 - 1] has two children where:

  • The left node has the value 2 * val, and
  • The right node has the value 2 * val + 1.

You are also given a 2D integer array queries of length m, where queries[i] = [ai, bi]. For each query, solve the following problem:

  1. Add an edge between the nodes with values ai and bi.
  2. Find the length of the cycle in the graph.
  3. Remove the added edge between nodes with values ai and bi.

Note that:

  • A cycle is a path that starts and ends at the same node, and each edge in the path is visited only once.
  • The length of a cycle is the number of edges visited in the cycle.
  • There could be multiple edges between two nodes in the tree after adding the edge of the query.

Return an array answer of length m where answer[i] is the answer to the ith query.

Example 1:

Input: n = 3, queries = [[5,3],[4,7],[2,3]]
Output: [4,5,3]
Explanation: The diagrams above show the tree of 23 - 1 nodes. Nodes colored in red describe the nodes in the cycle after adding the edge.
- After adding the edge between nodes 3 and 5, the graph contains a cycle of nodes [5,2,1,3]. Thus answer to the first query is 4. We delete the added edge and process the next query.
- After adding the edge between nodes 4 and 7, the graph contains a cycle of nodes [4,2,1,3,7]. Thus answer to the second query is 5. We delete the added edge and process the next query.
- After adding the edge between nodes 2 and 3, the graph contains a cycle of nodes [2,1,3]. Thus answer to the third query is 3. We delete the added edge.

Example 2:

Input: n = 2, queries = [[1,2]]
Output: [2]
Explanation: The diagram above shows the tree of 22 - 1 nodes. Nodes colored in red describe the nodes in the cycle after adding the edge.
- After adding the edge between nodes 1 and 2, the graph contains a cycle of nodes [2,1]. Thus answer for the first query is 2. We delete the added edge.

Constraints:

  • 2 <= n <= 30
  • m == queries.length
  • 1 <= m <= 105
  • queries[i].length == 2
  • 1 <= ai, bi <= 2n - 1
  • ai != bi
Patterns Used
🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an integer n. There is a complete binary tree with 2n - 1 nodes. The root of that tree is the node with the value 1, and every node with a value val in the range [1, 2n - 1 - 1] has two children where: The left node has the value 2 * val, and The right node has the value 2 * val + 1. You are also given a 2D integer array queries of length m, where queries[i] = [ai, bi]. For each query, solve the following problem: Add an edge between the nodes with values ai and bi. Find the length of the cycle in the graph. Remove the added edge between nodes with values ai and bi. Note that: A cycle is a path that starts and ends at the same node, and each edge in the path is visited only once. The length of a cycle is the number of edges visited in the cycle. There could be multiple edges between two nodes in the tree after adding the edge of the query. Return an array answer of length

Baseline thinking

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

Pattern signal: Array · Tree

Example 1

3
[[5,3],[4,7],[2,3]]

Example 2

2
[[1,2]]

Related Problems

  • Populating Next Right Pointers in Each Node (populating-next-right-pointers-in-each-node)
  • Lowest Common Ancestor of a Binary Tree (lowest-common-ancestor-of-a-binary-tree)
  • Path In Zigzag Labelled Binary Tree (path-in-zigzag-labelled-binary-tree)
Step 02

Core Insight

What unlocks the optimal approach

  • Find the distance between nodes “a” and “b”.
  • distance(a, b) = depth(a) + depth(b) - 2 * depth(LCA(a, b)). Where depth(a) denotes depth from root to node “a” and LCA(a, b) denotes the lowest common ancestor of nodes “a” and “b”.
  • To find LCA(a, b), iterate over all ancestors of node “a” and check if it is the ancestor of node “b” too. If so, take the one with maximum depth.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #2509: Cycle Length Queries in a Tree
class Solution {
    public int[] cycleLengthQueries(int n, int[][] queries) {
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int a = queries[i][0], b = queries[i][1];
            int t = 1;
            while (a != b) {
                if (a > b) {
                    a >>= 1;
                } else {
                    b >>= 1;
                }
                ++t;
            }
            ans[i] = t;
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.

Related Problems
#105Construct Binary Tree from Preorder and Inorder Traversalmedium→#106Construct Binary Tree from Inorder and Postorder Traversalmedium→#108Convert Sorted Array to Binary Search Treeeasy→#427Construct Quad Treemedium→#654Maximum Binary Treemedium→