leetcode/grind:50 subtree of another tree-brute force recursive

Author: kumar91gopi

README.md

@@ -338,7 +338,7 @@ solution of many challenges of [Leetcode](https://leetcode.com/), [Exercism](htt
 47. [Min Stack](https://github.com/kumar91gopi/Algorithms-and-Data-Structures-in-Ruby/blob/master/leetcode/grind75/min_stack.rb)
 48. [Balanced Binary tree](https://github.com/kumar91gopi/Algorithms-and-Data-Structures-in-Ruby/blob/master/leetcode/grind75/balanced_binary_tree.rb)
 49. [Symmetric Tree](https://github.com/kumar91gopi/Algorithms-and-Data-Structures-in-Ruby/blob/master/leetcode/grind75/symmetric_tree.rb)
-
+50. [Subtree of Another Tree](https://github.com/kumar91gopi/Algorithms-and-Data-Structures-in-Ruby/blob/master/leetcode/grind75/subtree_of_another_tree.rb)
 
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leetcode/grind75/subtree_of_another_tree.rb

@@ -0,0 +1,37 @@
+# Problem: https://leetcode.com/problems/subtree-of-another-tree/description/
+=begin
+Given the roots of two binary trees root and subRoot,
+return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.
+A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node's descendants.
+The tree tree could also be considered as a subtree of itself.
+=end
+
+# Definition for a binary tree node.
+# class TreeNode
+#     attr_accessor :val, :left, :right
+#     def initialize(val = 0, left = nil, right = nil)
+#         @val = val
+#         @left = left
+#         @right = right
+#     end
+# end
+# @param {TreeNode} root
+# @param {TreeNode} sub_root
+# @return {Boolean}
+
+# Solution 1 (Brute force recursive)
+def is_subtree(root, sub_root)
+    return true if sub_root.nil?
+    return false if root.nil?
+    is_found = false
+    if root.val == sub_root.val
+        return true if is_same_tree(root, sub_root)
+    end
+    is_subtree(root.left, sub_root) || is_subtree(root.right, sub_root)
+end
+
+def is_same_tree(root1, root2)
+    return true if root1.nil? and root2.nil?
+    return false if root1.nil? or root2.nil?
+    root1.val == root2.val and is_same_tree(root1.left, root2.left) and is_same_tree(root1.right, root2.right)
+end