# 1650. Lowest Common Ancestor of a Binary Tree III

Given two nodes of a binary tree `p` and `q`, return their lowest common ancestor (LCA).

Each node will have a reference to its parent node. The definition for `Node` is below:

```class Node {
public int val;
public Node left;
public Node right;
public Node parent;
}
```

According to the definition of LCA on Wikipedia: “The lowest common ancestor of two nodes p and q in a tree T is the lowest node 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 = 1
Output: 3
Explanation: 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 = 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.
```

Example 3:

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

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` exist in the tree.

Solution:

The last line works because `None` in python evaluates to `False`, so doing `None or Node` will return `Node`

``````class Solution(object):
def lowestCommonAncestor(self, p, q):
pVals = set()
def traverse_up(root):
if root == None or root in pVals:
return root
return traverse_up(root.parent)

return traverse_up(p) or traverse_up(q)``````

Solution2:

The idea is to store the parents (path) from root to p, and then check q’s path.

``````class Solution:
def lowestCommonAncestor(self, p, q):
path = set()
while p: