Trees
Practice & Revision
Binary Tree pattern recognition decision tree, cheat sheet, and Top 15 must-solve tree interview problems.
1. Introduction
This section serves as your comprehensive reference and practice guide for Trees. Master these templates, review the decision tree, and solve the curated Top 15 interview problems to prepare for technical interviews.
2. Binary Tree Pattern Decision Tree
Use this guide to identify the correct recursive or iterative strategy based on your problem:
3. Revision Cheat Sheet
Common DFS/BFS Templates Reference
| Pattern / Traversal | Processing Invariant | Code Template |
|---|---|---|
| Pre-Order DFS | Root first | process(node); dfs(node.left); dfs(node.right); |
| In-Order DFS | Left subtree first | dfs(node.left); process(node); dfs(node.right); |
| Post-Order DFS | Children first | dfs(node.left); dfs(node.right); process(node); |
| Level-Order BFS | Layer-by-layer | int len=q.size(); for(0..len){ curr=q.poll(); q.add(child); } |
| Height Checker | Bottom-up | if(node==null) return 0; return 1 + max(left, right); |
4. Top 15 Must-Solve Tree Problems
5. Problem-Solving Framework
When coding Binary Tree solutions, follow this checklist:
1. Verify the Base Cases:
- Write out recursion boundaries (if (root == null)) first.
2. Choose Top-Down vs Bottom-Up:
- Top-Down: Pass state parameters (like target sums or parent range boundaries) from root down to children.
- Bottom-Up: Compute heights or target nodes returned from left/right subtrees and combine them at the parent node.
3. Trace Skewed Configurations:
- Ensure your code does not crash and handles O(N) depth stack execution when nodes only branch in one direction.
6. Quiz
Question 1: In 'Validate Binary Search Tree', why does 'root.left.val < root.val' check fail to prove validity?
Answer: Because BST rules apply globally. A node's left child might be smaller than it, but that child's right descendant could be larger than the root node, violating the global BST property. We must pass bounding values down recursively.Question 2: What is the benefit of iterative BST search over recursive search?
Answer: Iterative BST search takesO(1) space. Because we only traverse down one branch without backtracking, we can use a simple while loop (curr = curr.left or curr = curr.right) without pushing call frames onto the stack.
Question 3: In C++, how does recursive post-order height search prevent memory leaks?
Answer: If we need to delete a tree, a post-order traversal ensures we recursively delete child subtrees first before callingdelete on the parent node, preventing orphans.
Question 4: What is the time complexity of 'Symmetric Tree' validation?
Answer:O(N) time, as we compare corresponding mirrors in both subtrees, visiting each node at most once.