# 235. 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.

```cpp
// Recursion
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { // time: O(n); space: O(n)
    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 root;
}
```

```cpp
// Iteration
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { // time: O(n); space: O(1)
    TreeNode* cur = root;
    while (cur) {
        if (cur->val > p->val && cur->val > q->val) 
            cur = cur->left;
        else if (cur->val < p->val && cur->val < q->val)
            cur = cur->right;
        else
            break;
    }
    return cur;
}
```