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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}deflowest_common_ancestor(root,p,q)returnnilifroot.nil?returnrootifroot.val == p.valorroot.val == q.valret_l=lowest_common_ancestor(root.left,p,q)ret_r=lowest_common_ancestor(root.right,p,q)ifret_l.nil?andret_r.nil?returnnilelsifret_l.nil?returnret_relsifret_r.nil?returnret_lelsereturnrootendend
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