LEETCODE 1650:C SOLUTION:Finding the Lowest Common Ancestor of a Binary Tree's Nodes

Welcome to this comprehensive guide on solving LeetCode Problem 1650, where we will dive into finding the Lowest Common Ancestor (LCA) of a binary tree's nodes using parent pointers. This video is designed to help you understand the problem statement, approach, and thought process required to solve this problem effectively. Whether you are preparing for coding interviews or looking to strengthen your problem-solving skills, this tutorial will provide you with valuable insights and techniques.
Introduction to LeetCode Problem 1650
LeetCode Problem 1650 challenges us to find the lowest common ancestor of two nodes in a binary tree where each node has a parent pointer. The parent pointer allows us to traverse from a node to its parent, which is crucial for solving this problem efficiently. Understanding the concept of the lowest common ancestor is essential as it frequently appears in coding interviews and real-world applications.
Problem Statement
Given a binary tree where each node has a parent pointer, we are asked to find the lowest common ancestor of two given nodes, p and q. The lowest common ancestor of two nodes p and q is defined as the deepest node that has both p and q as descendants. In this problem, each node has a parent pointer that points to its parent, and we need to leverage this information to find the solution.
Understanding the Tree Structure
Before we delve into the solution, it is crucial to understand the tree structure and how parent pointers work. In a typical binary tree, each node points to its left and right children. However, in this problem, each node also has a pointer to its parent, enabling upward traversal in the tree. This additional pointer simplifies the process of finding the LCA by allowing us to move from a node to its ancestors.
Approach to Solve the Problem
The approach to solving this problem involves traversing the tree from both nodes p and q simultaneously until we find their lowest common ancestor. We use a technique similar to finding the intersection point of two linked lists. The key steps are as follows:
Initialize Two Pointers: Start by initializing two pointers, a and b, to the nodes p and q, respectively.
Traverse Upwards: Move both pointers upwards in the tree by following their parent pointers.
Handle Null Pointers: If a pointer reaches the root (i.e., its parent is null), reset it to the other node (i.e., if a is null, set it to q, and if b is null, set it to p).
Find the LCA: Continue this process until the two pointers meet. The meeting point is the lowest common ancestor of nodes p and q.
Detailed Explanation
Initialization: We start by setting two pointers, a and b, to nodes p and q. These pointers will help us traverse the tree upwards from the given nodes.
Traversal: In each iteration of the loop, we move both pointers to their respective parents. If a pointer reaches the root (i.e., becomes null), we reset it to the other node. This reset ensures that both pointers traverse equal distances, covering both subtrees and eventually meeting at the LCA.
Meeting Point: When the two pointers meet, they point to the lowest common ancestor. This is because resetting the pointers allows them to traverse the same number of steps, ensuring they meet at the deepest common ancestor.
Edge Cases
Nodes are the Same: If p and q are the same node, the LCA is the node itself.
One Node is Ancestor of the Other: If one node is an ancestor of the other, the LCA is the ancestor node.
Tree is Rooted at a Single Node: If the tree consists of a single node, the LCA of any two nodes (which would be the same node) is the node itself.
Time Complexity
The time complexity of this solution is O(h), where h is the height of the tree. This is because each pointer traverses the tree upwards, and the maximum distance they can travel is the height of the tree.
Space Complexity
The space complexity is O(1) as we are only using a fixed amount of extra space for the pointers a and b.
Conclusion
In this video, we explored an efficient solution to LeetCode Problem 1650 by leveraging parent pointers in a binary tree to find the lowest common ancestor of two given nodes. We discussed the problem statement, tree structure, and approach in detail, along with edge cases and complexity analysis. Understanding this technique is not only beneficial for coding interviews but also for solving real-world problems involving tree data structures.
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