Subtree of Another Tree
Detailed guide and Python implementation for the 'Subtree of Another Tree' problem.
1. Concept Overview
The 'Subtree of Another Tree' problem is a key challenge in the Trees section.
This implementation focuses on medium-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Subtree of Another Tree.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Subtree of Another Tree carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.
A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node's descendants. The tree tree could also be considered as a subtree of itself.
The trees are represented as level-order lists. Implement a function isSubtree(root: list, subRoot: list) -> bool.
- •The number of nodes in the root tree is in the range [1, 2000]
- •The number of nodes in the subRoot tree is in the range [1, 1000]
- •-10000 <= root.val <= 10000
- •-10000 <= subRoot.val <= 10000
Examples
[3,4,5,1,2], [4,1,2]
True
The subtree rooted at node 4 in the main tree matches subRoot exactly.
[3,4,5,1,2,None,None,None,None,0], [4,1,2]
False
The subtree rooted at 4 in the main tree has an extra node 0 under 2, so it doesn't match subRoot.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def is_subtree_opt(root, subRoot):
if isinstance(root, list):
r1 = build_tree(root)
r2 = build_tree(subRoot)
return is_subtree_opt_helper(r1, r2)
return is_subtree_opt_helper(root, subRoot)
def is_subtree_opt_helper(root: TreeNode, subRoot: TreeNode) -> bool:
def is_same(p, q):
if not p and not q:
return True
if not p or not q or p.val != q.val:
return False
return is_same(p.left, q.left) and is_same(p.right, q.right)
if not subRoot: return True
if not root: return False
if is_same(root, subRoot): return True
return is_subtree_opt_helper(root.left, subRoot) or is_subtree_opt_helper(root.right, subRoot)Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def is_subtree_brute(root, subRoot):
if isinstance(root, list):
r1 = build_tree(root)
r2 = build_tree(subRoot)
return is_subtree_brute_helper(r1, r2)
return is_subtree_brute_helper(root, subRoot)
def is_subtree_brute_helper(root: TreeNode, subRoot: TreeNode) -> bool:
def same(p, q):
if not p and not q: return True
if not p or not q or p.val != q.val: return False
return same(p.left, q.left) and same(p.right, q.right)
if not root: return False
if same(root, subRoot): return True
return is_subtree_brute_helper(root.left, subRoot) or is_subtree_brute_helper(root.right, subRoot)Algorithm Pattern Checklist
When dealing with Trees data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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