# Question

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).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

```
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4
```

Example 1:

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

Note:

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

# Solution

The question becomes finding the smallest subtree that contains both p and q and returning its root. A subtree contains a node is either a node in its left tree, right tree, or its root. Because of the recursive call, we can store the first occurrence of such tree when both p and q exists, and it is the LCA.

```
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
self.lca = None
self.contains(root, p, q)
return self.lca
def contains(self, root, p, q):
if root is None:
return False, False
left_p, left_q = self.contains(root.left, p, q)
right_p, right_q = self.contains(root.right, p, q)
if self.lca is not None:
return True, True
contains_p = root.val == p.val or left_p or right_p
contains_q = root.val == q.val or left_q or right_q
if contains_p and contains_q:
self.lca = root
return contains_p, contains_q
```