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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).”

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.

Solutions (Ruby)

1. Recursion

# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val)
#         @val = val
#         @left, @right = nil, nil
#     end
# end

# @param {TreeNode} root
# @param {TreeNode} p
# @param {TreeNode} q
# @return {TreeNode}
def lowest_common_ancestor(root, p, q)
    return nil if root.nil?
    return root if root.val == p.val or root.val == q.val

    ret_l = lowest_common_ancestor(root.left, p, q)
    ret_r = lowest_common_ancestor(root.right, p, q)

    if ret_l.nil? and ret_r.nil?
        return nil
    elsif ret_l.nil?
        return ret_r
    elsif ret_r.nil?
        return ret_l
    else
        return root
    end
end