236. 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).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,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 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 in python:

# 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':
        hashset = {}
        def record(node):
            if  node.left:
                hashset[node.left.val] = node
                record(node.left)
            if  node.right:
                hashset[node.right.val] = node
                record(node.right)
        record(root)
        hashp = []
        parent = p
        while parent.val != root.val:
            hashp.append(parent.val)
            parent = hashset[parent.val]
        hashp.append(root.val)
        parent = q
        while parent.val not in hashp:
            parent = hashset[parent.val]
        return parent

Complexity analysis:

• Time complexity: O(N)
• Space complexity: O(N)