The problem titled “Find a Corresponding Node of a Binary Tree in a Clone of That Tree” is a fascinating challenge frequently seen on Leetcode. This problem involves binary tree traversal and testing your understanding of Python’s object-oriented features. In this comprehensive guide, we’ll discuss the problem in detail, explore multiple solutions, and provide tips to achieve optimal performance.
Table of Contents
- Understanding the Problem Statement
- Introducing the Binary Tree Data Structure
- Initial Thoughts and Pseudocode
- Solving Using Depth-First Search (DFS)
- Solving Using Breadth-First Search (BFS)
- Time and Space Complexity Analysis
- Testing and Edge Cases
1. Understanding the Problem Statement
In this problem, you are given two binary trees,
cloned, and a reference to a node
target in the
original tree. The
cloned tree is a copy of the
original tree. Your task is to return a reference to the same node in the
Constraints and Assumptions
- The number of nodes in the tree is in the range
- The values of the nodes of the
originaltree are unique.
targetnode is a reference to a node in the
clonedis a copy of
2. Introducing the Binary Tree Data Structure
Before diving into the problem, let’s define the TreeNode class, which we will use to represent the nodes in the tree.
class TreeNode: def __init__(self, value=0, left=None, right=None): self.val = value self.left = left self.right = right
3. Initial Thoughts and Pseudocode
We can solve this problem by traversing both trees simultaneously until we find the node in the
original tree that matches the
target. Once we do, we’ll return the corresponding node in the
Pseudocode for DFS approach:
1. Start DFS traversal from the root of the `original` and `cloned` trees. 2. If you find `target` in `original`, return the corresponding node in `cloned`. 3. Otherwise, traverse the left and right subtrees.
4. Solving Using Depth-First Search (DFS)
The following Python code solves the problem using DFS.
def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode: if not original: return None if original == target: return cloned left = self.getTargetCopy(original.left, cloned.left, target) if left: return left right = self.getTargetCopy(original.right, cloned.right, target) if right: return right
5. Solving Using Breadth-First Search (BFS)
You can also solve this problem using BFS, where you’d use a queue to traverse the tree level by level.
from collections import deque def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode: queue = deque([(original, cloned)]) while queue: ori, clo = queue.popleft() if ori == target: return clo if ori.left: queue.append((ori.left, clo.left)) if ori.right: queue.append((ori.right, clo.right))
6. Time and Space Complexity Analysis
- DFS and BFS both have a time complexity of O(n), where n is the number of nodes in the tree.
- The space complexity is also O(n) for both methods due to the maximum recursion depth for DFS and the maximum queue size for BFS.
7. Testing and Edge Cases
Test your solution on multiple test cases to make sure it handles all scenarios, including edge cases like a single-node tree or a completely unbalanced tree.
The problem “Find a Corresponding Node of a Binary Tree in a Clone of That Tree” is an excellent test of your understanding of tree traversal algorithms and object-oriented programming in Python. Both DFS and BFS are viable solutions, with comparable time and space complexity.