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

**Note:**

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

## Solution & Analysis

### Using attribution of a 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) {

        // Value of current node or parent node.
        int parentVal = root.val;

        // Value of p
        int pVal = p.val;

        // Value of q;
        int qVal = q.val;

        if (pVal > parentVal && qVal > parentVal) {
            // If both p and q are greater than parent
            return lowestCommonAncestor(root.right, p, q);
        } else if (pVal < parentVal && qVal < parentVal) {
            // If both p and q are lesser than parent
            return lowestCommonAncestor(root.left, p, q);
        } else {
            // We have found the split point, i.e. the LCA node.
            return root;
        }
    }
}
```

Iterative

```java
```

### Same code as Lowest Common Ancestor of a Binary Tree

```java
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root == q) return root;
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if(left != null && right != null) return root;
        return left != null ? left : right;
    }
}
```

## Reference

<https://www.lintcode.com/problem/lowest-common-ancestor-ii/description>

<https://www.lintcode.com/problem/lowest-common-ancestor-iii/description>


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