# LeetCode 236. Lowest Common Ancestor of a Binary Tree

## 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 = 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 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).

## One Thought to “LeetCode 236. Lowest Common Ancestor of a Binary Tree”

1. Frank says:

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