> For the complete documentation index, see [llms.txt](https://shuati.gitbook.io/crack-lintcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://shuati.gitbook.io/crack-lintcode/linkedin/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]

```
        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5
```

**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.
```

**Note:**

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

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null)
            return null;
        
        if(root.val > Math.max(p.val,q.val)){
            return lowestCommonAncestor(root.left,p,q);
        }
        
        if(root.val < Math.min(p.val,q.val)){
            return lowestCommonAncestor(root.right,p,q);
        }
        
       else 
           return root;
    }
}
```