Step-By-Step Directions From a Binary Tree Node to Another
Time O(n) · Space O(h) · Official statement on LeetCode
Solutions
// Time: O(n)
// Space: O(h)
class Solution {
public:
string getDirections(TreeNode* root, int startValue, int destValue) {
auto src = iter_dfs(root, startValue);
auto dst = iter_dfs(root, destValue);
while (!empty(src) && !empty(dst) && src.back() == dst.back()) {
src.pop_back();
dst.pop_back();
}
reverse(begin(dst), end(dst));
return string(size(src), 'U') + dst;
}
private:
string iter_dfs(TreeNode *root, int val) {
string path;
vector<tuple<int, TreeNode *, char>> stk = {{1, root, ' '}};
while (!empty(stk)) {
auto [step, node, d] = stk.back(); stk.pop_back();
if (step == 1) {
if (node->val == val) {
reverse(begin(path), end(path));
return path;
}
for (const auto& child : {node->left, node->right}) {
if (!child) {
continue;
}
stk.emplace_back(3, nullptr, ' ');
stk.emplace_back(1, child, ' ');
stk.emplace_back(2, nullptr, (child == node->left) ? 'L' : 'R');
}
} else if (step == 2) {
path.push_back(d);
} else if (step == 3) {
path.pop_back();
}
}
return "";
}
};
// Time: O(n)
// Space: O(h)
class Solution2 {
public:
string getDirections(TreeNode* root, int startValue, int destValue) {
string src, dst;
dfs(root, startValue, &src);
dfs(root, destValue, &dst);
while (!empty(src) && !empty(dst) && src.back() == dst.back()) {
src.pop_back();
dst.pop_back();
}
reverse(begin(dst), end(dst));
return string(size(src), 'U') + dst;
}
private:
bool dfs(TreeNode *curr, int val, string *path) {
if (curr->val == val) {
return true;
}
if (curr->left && dfs(curr->left, val, path)) {
path->push_back('L');
} else if (curr->right && dfs(curr->right, val, path)) {
path->push_back('R');
}
return !empty(*path);
}
};
Beginner Explanation
What is Step-By-Step Directions From a Binary Tree Node to Another?
Step-By-Step Directions From a Binary Tree Node to Another (LeetCode #2096) is a Medium problem that primarily trains tree.
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 dfs backtracking and stack.
- 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: DFS, Stack.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Step-By-Step Directions From a Binary Tree Node to Another
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 dfs backtracking and stack.
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(n)) and space (O(h)) 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(n) time and O(h) space.
Pattern focus: dfs backtracking and stack
Use the pattern as a checklist:
- dfs backtracking — confirm the invariant holds after each step
- stack — 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(n) |
| Space | O(h) |
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 Step-By-Step Directions From a Binary Tree Node to Another
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dfs backtracking and stack — 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 dfs backtracking and stack:
- 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: tree.
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 Step-By-Step Directions From a Binary Tree Node to Another in a second language (cpp, python).
- Drill 3–5 more problems tagged tree.
- 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 dfs backtracking and stack 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
Step-By-Step Directions From a Binary Tree Node to Another (#2096) — Medium. Pattern: dfs backtracking and stack. Complexity: O(n) time / O(h) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Step-By-Step Directions From a Binary Tree Node to Another?+
The reference solutions aim for O(n) time and O(h) space. Always re-derive complexity from the code you write in the interview.
What pattern does Step-By-Step Directions From a Binary Tree Node to Another use?+
It primarily maps to dfs backtracking and stack, within the broader topic of tree.
Is Step-By-Step Directions From a Binary Tree Node to Another 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/step-by-step-directions-from-a-binary-tree-node-to-another/