Tree Diameter
Time O(\ · Space V\ · Official statement on LeetCode
Solutions
// Time: O(|V| + |E|)
// Space: O(|E|)
// iterative dfs
class Solution {
public:
int treeDiameter(vector<vector<int>>& edges) {
vector<vector<int>> adj(size(edges) + 1);
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
int result = 0;
const auto& iter_dfs = [&]() {
int result = 0;
using RET = int;
RET ret{};
vector<tuple<int, int, int, shared_ptr<RET>, RET *>> stk = {{1, 0, -1, nullptr, &ret}};
while (!empty(stk)) {
const auto [step, u, p, ret2, ret] = stk.back(); stk.pop_back();
if (step == 1) {
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
const auto& ret2 = make_shared<RET>();
stk.emplace_back(2, -2, -2, ret2, ret);
stk.emplace_back(1, v, u, nullptr, ret2.get());
}
} else if (step == 2) {
result = max(result, *ret + (*ret2 + 1));
*ret = max(*ret, *ret2 + 1);
}
}
return result;
};
return iter_dfs();
}
};
// Time: O(|V| + |E|)
// Space: O(|E|)
// dfs
class Solution2 {
public:
int treeDiameter(vector<vector<int>>& edges) {
vector<vector<int>> adj(size(edges) + 1);
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
int result = 0;
const function<int (int, int)> dfs = [&](int u, int p) {
int mx = 0;
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
const int curr = dfs(v, u);
result = max(result, mx + (curr + 1));
mx = max(mx, curr + 1);
}
return mx;
};
dfs(0, -1);
return result;
}
};
// Time: O(|V| + |E|)
// Space: O(|E|)
// bfs, tree dp
class Solution3 {
public:
int treeDiameter(vector<vector<int>>& edges) {
vector<vector<int>> adj(size(edges) + 1);
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
int result = 0;
const auto& bfs = [&]() {
int result = 0;
vector<int> dp(size(adj));
vector<int> degree(size(adj));
vector<int> q;
for (int u = 0; u < size(adj); ++u) {
degree[u] = size(adj[u]);
if (degree[u] == 1) {
q.emplace_back(u);
}
}
while (!empty(q)) {
vector<int> new_q;
for (const auto& u : q) {
if (degree[u] == 0) {
continue;
}
--degree[u];
for (const auto& v : adj[u]) {
if (degree[v] == 0) {
continue;
}
result = max(result, dp[v] + (dp[u] + 1));
dp[v] = max(dp[v], dp[u] + 1);
if (--degree[v] == 1) {
new_q.emplace_back(v);
}
}
}
q = move(new_q);
}
return result;
};
return bfs();
}
};
// Time: O(|V| + |E|)
// Space: O(|E|)
// bfs
class Solution4 {
public:
int treeDiameter(vector<vector<int>>& edges) {
vector<vector<int>> adj(size(edges) + 1);
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
int result = 0;
const auto& bfs = [&](int root) {
int d = -1, new_root = -1;
vector<bool> lookup(size(adj));
lookup[root] = true;
vector<int> q = {root};
while (!empty(q)) {
++d;
new_root = q[0];
vector<int> new_q;
for (const auto& u : q) {
for (const auto& v : adj[u]) {
if (lookup[v]) {
continue;
}
lookup[v] = true;
new_q.emplace_back(v);
}
}
q = move(new_q);
}
return pair(d, new_root);
};
const auto& [_, root] = bfs(0);
const auto& [d, __] = bfs(root);
return d;
}
};
Beginner Explanation
What is Tree Diameter?
Tree Diameter (LeetCode #1245) is a Medium problem that primarily trains breadth first search.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with queue bfs.
- Only then translate the idea into code.
Why this problem matters
It sits in the sweet spot of interview difficulty: multiple valid approaches, clear trade-offs. Official solution notes mention: )_.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Tree Diameter
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to queue bfs.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O() and space (V) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target *O(* time and *V* space.
Pattern focus: queue bfs
Use the pattern as a checklist:
- queue bfs — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | *O(* |
| Space | *V* |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Tree Diameter
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for queue bfs — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to queue bfs:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: breadth first search.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Tree Diameter in a second language (cpp, python).
- Drill 3–5 more problems tagged breadth first search.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the queue bfs approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Tree Diameter (#1245) — Medium. Pattern: queue bfs. Complexity: O(\ time / V\ space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Tree Diameter?+
The reference solutions aim for O(\ time and V\ space. Always re-derive complexity from the code you write in the interview.
What pattern does Tree Diameter use?+
It primarily maps to queue bfs, within the broader topic of breadth first search.
Is Tree Diameter good for interviews?+
Yes — as a Medium problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/tree-diameter/