## 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 = 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`

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).