# Lowest Common Ancestor of a Binary Search Tree LeetCode Solution Share:

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: “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, 105]`.
• `-109 <= Node.val <= 109`
• All `Node.val` are unique.
• `p != q`
• `p` and `q` will exist in the BST.

Time: O(h)
Space: O(h)

### C++

``````class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (root->val > max(p->val, q->val))
return lowestCommonAncestor(root->left, p, q);
if (root->val < min(p->val, q->val))
return lowestCommonAncestor(root->right, p, q);
return root;
}
};
``````

### Java

`````` class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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);
return root;
}
}

``````

### Python

``````
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if root.val > max(p.val, q.val):
return self.lowestCommonAncestor(root.left, p, q)
if root.val < min(p.val, q.val):
return self.lowestCommonAncestor(root.right, p, q)
return root
``````

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