# LeetCode 1650. Lowest Common Ancestor of a Binary Tree III

## Description

https://leetcode.com/problems/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.

## Explanation

The problem interface is different from Lowest Common Ancestor of a Binary Tree. Find the root of the tree, and use the same approach to find the lowest common ancestor between two nodes.

## Python Solution

``````"""
# Definition for a Node.
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
self.parent = None
"""

class Solution(object):
def lowestCommonAncestor(self, p, q):
"""
:type node: Node
:rtype: Node
"""

root = p

while root.parent != None:
root = root.parent

print (root.val)
return self.helper(root, p, q)

def helper(self, root, p, q):
if not root:
return None

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

return None

``````
• Time Complexity: O(N).
• Space Complexity: O(N).