Hello, coding test applicants! Today we will delve deep into the problem of Lowest Common Ancestor (LCA). This problem is related to tree structures and requires a deep understanding of algorithms and data structures. We will also explore how to solve this problem using C#.
Problem Definition
The problem is to find the lowest common ancestor of two nodes in a given binary tree. Given two nodes A and B, our goal is to find the deepest node that is an ancestor of both A and B.
Example
For instance, let's assume there is a binary tree as follows. 3 / \ 5 1 / \ / \ 6 2 0 8 / \ 7 4 The lowest common ancestor of nodes 5 and 1 is 3. The lowest common ancestor of nodes 5 and 2 is 5. The lowest common ancestor of nodes 6 and 4 is 5.
Approach to the Problem
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Understanding Tree Structure
First, it is essential to understand the structure of a binary tree. Each node can have at most two children, and it can be divided into a left subtree and a right subtree. You should know how to traverse the tree to find nodes.
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Recursive Approach
One of the main approaches is to search recursively. If the current node is either A or B, return the current node and explore the left and right subtrees to find the other node. If both are found, then the current node is the lowest common ancestor.
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Writing Tree Traversal Algorithm
Now, we need to write the code in C# to implement the algorithm. In this process, we will incorporate recursion to navigate and verify nodes through the code.
C# Code Implementation
// Definition of binary tree node class public class TreeNode { public int Value; public TreeNode Left; public TreeNode Right; public TreeNode(int value) { Value = value; Left = null; Right = null; } } public class Solution { public TreeNode LowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { // Base case: If root is null or root is p or q, return root if (root == null || root == p || root == q) { return root; } // Search each node in the left and right subtrees TreeNode left = LowestCommonAncestor(root.Left, p, q); TreeNode right = LowestCommonAncestor(root.Right, p, q); // If p and q are in different subtrees, the current node is LCA if (left != null && right != null) { return root; } // If only one is found, return that node return left ?? right; } }
Code Explanation
This code is written to find the lowest common ancestor of two nodes in a given binary tree. Below are the descriptions of the key parts of the code:
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TreeNode Class
This class represents each node in the tree. It stores the value of each node, the left child node, and the right child node.
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LowestCommonAncestor Method
This method is called recursively and checks if the current node is null or if it is p or q. It searches for p and q in the left and right subtrees, and if both are found in different subtrees, it returns the current node.
Complexity Analysis
The time complexity of this algorithm is O(N), where N represents the number of nodes in the tree. It is proportional to the need to traverse all nodes. The space complexity is O(H), where H represents the height of the tree. This corresponds to the depth of the recursion call stack.
Example Execution
// Example tree creation and finding LCA TreeNode root = new TreeNode(3); root.Left = new TreeNode(5); root.Right = new TreeNode(1); root.Left.Left = new TreeNode(6); root.Left.Right = new TreeNode(2); root.Right.Left = new TreeNode(0); root.Right.Right = new TreeNode(8); root.Left.Right.Left = new TreeNode(7); root.Left.Right.Right = new TreeNode(4); Solution solution = new Solution(); TreeNode p = root.Left; // Node 5 TreeNode q = root.Left.Right; // Node 2 TreeNode lca = solution.LowestCommonAncestor(root, p, q); Console.WriteLine($"Lowest Common Ancestor: {lca.Value}"); // Outputs 5
Conclusion
Today, we explored how to solve the lowest common ancestor problem using C#. This problem serves as a foundation for tackling various algorithmic challenges and requires an understanding of tree structure and recursive thinking. Through practice, I hope you will enhance your coding skills and increase your understanding of C# during this opportunity.