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

According to the definition of LCA on Wikipedia: “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:**

```
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.
```

**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, 10`

.^{5}] `-10`

^{9}<= Node.val <= 10^{9}- All
`Node.val`

are**unique**. `p != q`

`p`

and`q`

will exist in the BST.

```
def lowestCommonAncestor(self, root, p, q):
while (root.val - p.val) * (root.val - q.val) > 0:
root = (root.left, root.right)[p.val > root.val]
return root
```

```
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
while ((root.val - p.val) * (root.val - q.val) > 0)
root = p.val < root.val ? root.left : root.right;
return root;
}
```

```
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if ((root -> val > p -> val) && (root -> val > q -> val)) {
return lowestCommonAncestor(root -> left, p, q);
}
if ((root -> val < p -> val) && (root -> val < q -> val)) {
return lowestCommonAncestor(root -> right, p, q);
}
return root;
}
};
```

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