## Lowest Common Ancestor of a Binary Search Tree Link

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).”

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

```
4
/ \
3 6
/ \ / \
0 8 5 7
/ \
1 9
```

#### Example:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 3, q = 6

Output: 7

Explanation: The LCA of nodes 6 and 3 is 7.

**Note**:

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

### Solution:

This is the problem given only to check your basic understanding of BST. First we go through the basic points on BST:

- Left subtree of a node N contains nodes whose values are lesser than or equal to node N’s value.
- Right subtree of a node N contains nodes whose values are greater than node N’s value.
- Both left and right subtrees are also BSTs.

These basic understanding is the logic behind it’s implementation:

### Implementation:

### Java:

```
public static class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || p == root || q == root) {
return root;
}
if ((root.val - p.val) * (root.val - q.val) > 0) {
if (root.val - p.val > 0) {
return lowestCommonAncestor(root.left, p, q);
}
return lowestCommonAncestor(root.right, p, q);
}
return root;
}
}
```

### Complexity Analysis:

**Time Complexity:**O(n).**Space Complexity:**O(n).