Lowest Common Ancestor of a Binary Search Tree

LeetCode 235 | Difficulty: Medium

Medium

Problem Description

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).”

Example 1:

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:

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, 10^5]`.

- `-10^9 <= Node.val <= 10^9`

- All `Node.val` are **unique**.

- `p != q`

- `p` and `q` will exist in the BST.

Topics: Tree, Depth-First Search, Binary Search Tree, Binary Tree

---

Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

---

Solutions

Solution 1: C# (Best: 156 ms)

Metric	Value
Runtime	156 ms
Memory	N/A
Date	2018-04-25

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

---

Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$

---

Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.