LC 235. Lowest Common Ancestor of a Binary Search Tree

Alen Alex · February 8, 2024

LeetCode Link: 235. Lowest Common Ancestor of a Binary Search Tree
Language: C#

Problem Statement

Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Examples

Example 1:

img1
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2

img1
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [2,1], p = 2, q = 1
Output: 2

Constraints

  • The number of nodes in the tree is in the range [2, 105].
  • -109 <= Node.val <= 109
  • All Node.val are unique.
  • p != q
  • p and q will exist in the BST.

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */

public class Solution 
{
    public TreeNode LowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) 
    {
        if (root is null)
        {
            return null;
        }

        if ((root.val >= p.val && root.val <= q.val) ||
            (root.val <= p.val && root.val >= q.val))
            {
                return root;
            }
        
        if (root.val > p.val && root.val > q.val)
        {
            return LowestCommonAncestor(root.left, p, q);
        }
        
        return LowestCommonAncestor(root.right, p, q);
    }
}

Complexity

Time Complexity: O(N)
Space Complexity: O(N)

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