## Description

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

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

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 = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1Output:3Explanation:The LCA of nodes 5 and 1 is 3.

**Example 2:**

Input:root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4Output:5Explanation:The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

**Example 3:**

Input:root = [1,2], p = 1, q = 2Output:1

**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 tree.

## Explanation

If p and q in the left and right subtrees, return root.

If in one side of the subtree, find the LCA in that subtree.

If the root is p or q or is null, return root.

```
# 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.lca_helper(root, p, q)
def lca_helper(self, root, p, q):
if not root:
return None
if p == root or q == root:
return root
left_lca = self.lca_helper(root.left, p, q)
right_lca = self.lca_helper(root.right, p, q)
if left_lca and right_lca:
return root
if left_lca:
return left_lca
if right_lca:
return right_lca
return None
```

- Time Complexity: O(N).
- Space Complexity: O(1).

Are you sure the space complexity is O(1) if you use recursion?