# 0235. Lowest Common Ancestor of a Binary Search Tree

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree

## Description

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**).”

**Example 1:**

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

```
**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:**

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

```
**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.

## ac

```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) {
        // exit
        if (root == null || root.val == p.val || root.val == q.val) return root;

        // p/q in different subtree
        if (p.val < root.val && root.val < q.val || q.val < root.val && root.val < p.val) {
            return root;
        } 
        // p/q in same subtree, recursion
        else if (p.val < root.val) {
            return lowestCommonAncestor(root.left, p, q);
        } else {
            return lowestCommonAncestor(root.right, p, q);
        }
    }
}

/*
1)p/q in different subtree, return root; 2) p/q in same subtree, recursion. if root.val == p.val || root.val == q.val return current root, root must be the answer
*/
```
