Clone Binary Tree With Random Pointer
Time O(n) · Space O(h) · Official statement on LeetCode
Solutions
// Time: O(n)
// Space: O(h)
/**
* Definition for a Node.
* struct Node {
* int val;
* Node *left;
* Node *right;
* Node *random;
* Node() : val(0), left(nullptr), right(nullptr), random(nullptr) {}
* Node(int x) : val(x), left(nullptr), right(nullptr), random(nullptr) {}
* Node(int x, Node *left, Node *right, Node *random) : val(x), left(left), right(right), random(random) {}
* };
*/
class Solution {
public:
NodeCopy* copyRandomBinaryTree(Node* root) {
const auto& merge = [](Node *node) {
auto copy = new NodeCopy(node->val);
tie(node->left, copy->left) = pair(reinterpret_cast<Node*>(copy),
reinterpret_cast<NodeCopy*>(node->left));
return pair{reinterpret_cast<Node*>(copy->left), copy};
};
const auto& clone = [](Node *node) {
auto copy = reinterpret_cast<NodeCopy*>(node->left);
node->left->random = node->random ? node->random->left : nullptr;
node->left->right = node->right ? node->right->left : nullptr;
return pair{reinterpret_cast<Node*>(copy->left), copy};
};
const auto& split = [](Node *node) {
auto copy = reinterpret_cast<NodeCopy*>(node->left);
tie(node->left, copy->left) = pair(copy->left ? reinterpret_cast<Node*>(copy->left) : nullptr,
copy->left ? copy->left->left : nullptr);
return pair{reinterpret_cast<Node*>(node->left), copy};
};
iter_dfs(root, merge);
iter_dfs(root, clone);
return iter_dfs(root, split);
}
private:
NodeCopy *iter_dfs(Node *root, const function<pair<Node*, NodeCopy*>(Node*)>& callback) {
NodeCopy *result = nullptr;
vector<Node *> stk = {root};
while (!stk.empty()) {
auto node = stk.back(); stk.pop_back();
if (!node) {
continue;
}
const auto& [left_node, copy] = callback(node);
if (!result) {
result = copy;
}
stk.emplace_back(node->right);
stk.emplace_back(left_node);
}
return result;
}
};
// Time: O(n)
// Space: O(h)
class Solution_Recu {
public:
NodeCopy* copyRandomBinaryTree(Node* root) {
const auto& merge = [](Node *node) {
auto copy = new NodeCopy(node->val);
tie(node->left, copy->left) = pair(reinterpret_cast<Node*>(copy),
reinterpret_cast<NodeCopy*>(node->left));
return pair{reinterpret_cast<Node*>(copy->left), copy};
};
const auto& clone = [](Node *node) {
auto copy = reinterpret_cast<NodeCopy*>(node->left);
node->left->random = node->random ? node->random->left : nullptr;
node->left->right = node->right ? node->right->left : nullptr;
return pair{reinterpret_cast<Node*>(copy->left), copy};
};
const auto& split = [](Node *node) {
auto copy = reinterpret_cast<NodeCopy*>(node->left);
tie(node->left, copy->left) = pair(copy->left ? reinterpret_cast<Node*>(copy->left) : nullptr,
copy->left ? copy->left->left : nullptr);
return pair{reinterpret_cast<Node*>(node->left), copy};
};
dfs(root, merge);
dfs(root, clone);
return dfs(root, split);
}
private:
NodeCopy *dfs(Node *node, const function<pair<Node*, NodeCopy*>(Node*)>& callback) {
if (!node) {
return nullptr;
}
const auto& [left_node, copy] = callback(node);
dfs(left_node, callback);
dfs(node->right, callback);
return copy;
}
};
// Time: O(n)
// Space: O(n)
class Solution2 {
public:
NodeCopy* copyRandomBinaryTree(Node* root) {
unordered_map<Node*, NodeCopy*> lookup;
lookup[nullptr] = nullptr;
vector<Node *> stk = {root};
while (!stk.empty()) {
auto node = stk.back(); stk.pop_back();
if (!node) {
continue;
}
if (!lookup.count(node)) {
lookup[node] = new NodeCopy();
}
if (!lookup.count(node->left)) {
lookup[node->left] = new NodeCopy();
}
if (!lookup.count(node->right)) {
lookup[node->right] = new NodeCopy();
}
if (!lookup.count(node->random)) {
lookup[node->random] = new NodeCopy();
}
lookup[node]->val = node->val;
lookup[node]->left = lookup[node->left];
lookup[node]->right = lookup[node->right];
lookup[node]->random = lookup[node->random];
stk.emplace_back(node->right);
stk.emplace_back(node->left);
}
return lookup[root];
}
};
// Time: O(n)
// Space: O(n)
class Solution2_Recu {
public:
NodeCopy* copyRandomBinaryTree(Node* root) {
unordered_map<Node*, NodeCopy*> lookup;
lookup[nullptr] = nullptr;
dfs(root, &lookup);
return lookup[root];
}
private:
void dfs(Node *node, unordered_map<Node*, NodeCopy*> *lookup) {
if (!node) {
return;
}
if (!lookup->count(node)) {
(*lookup)[node] = new NodeCopy();
}
if (!lookup->count(node->left)) {
(*lookup)[node->left] = new NodeCopy();
}
if (!lookup->count(node->right)) {
(*lookup)[node->right] = new NodeCopy();
}
if (!lookup->count(node->random)) {
(*lookup)[node->random] = new NodeCopy();
}
(*lookup)[node]->val = node->val;
(*lookup)[node]->left = (*lookup)[node->left];
(*lookup)[node]->right = (*lookup)[node->right];
(*lookup)[node]->random = (*lookup)[node->random];
dfs(node->left, lookup);
dfs(node->right, lookup);
}
};
Beginner Explanation
What is Clone Binary Tree With Random Pointer?
Clone Binary Tree With Random Pointer (LeetCode #1485) is a Medium problem that primarily trains depth 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 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 Clone Binary Tree With Random Pointer
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 Clone Binary Tree With Random Pointer
- 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: depth 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 Clone Binary Tree With Random Pointer in a second language (cpp, python).
- Drill 3–5 more problems tagged depth 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 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
Clone Binary Tree With Random Pointer (#1485) — 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 Clone Binary Tree With Random Pointer?+
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 Clone Binary Tree With Random Pointer use?+
It primarily maps to dfs backtracking and stack, within the broader topic of depth first search.
Is Clone Binary Tree With Random Pointer 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/clone-binary-tree-with-random-pointer/