## Description

https://leetcode.com/problems/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 = 8Output:6Explanation: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 = 4Output:2Explanation: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 = 1Output: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.

## Explanation

If root is either p or q, LCA is root.

If LCA is in either subtree, return the LCA from that subtree.

If p and q are in different subtree, return the root.

## Python Solution

```
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
return self.helper(root, p, q)
def helper(self, root, p, q):
if not root:
return root
if root == p or root == q:
return root
left = self.helper(root.left, p, q)
right = self.helper(root.right, p, q)
if left and right:
return root
if left:
return left
if right:
return right
```

- Time Complexity: O(N).
- Space Complexity: O(N).