> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/trees/lowest-common-ancestor-of-a-binary-search-tree.md).

# Lowest Common Ancestor of a Binary Search Tree

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

According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “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**).”

Given binary search tree:  root = \[6,2,8,0,4,7,9,null,null,3,5]

![](https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)

**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.
```

**Constraints:**

* All of the nodes' values will be unique.
* p and q are different and both values will exist in the BST.

```java
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null)
            return null;
        // If one node is on right and other on left, then this node is LCA
        if ((root.val > p.val && root.val < q.val) || (root.val < p.val && root.val > q.val))
            return root;
        // If we have not found a splitting node(LCA), then this node will be considered LCA
        if (root.val == p.val || root.val == q.val)
            return root;
        // If both nodes are greater that current node, we will search on right
        if (p.val > root.val && q.val > root.val)
            return lowestCommonAncestor(root.right, p, q);
        // Else we will search on left
        return lowestCommonAncestor(root.left, p, q);
    }
}
```