Water - Kareha

lib/tree.rb

@@ -16,51 +16,199 @@ def initialize
     @root = nil
   end
 
-  # Time Complexity: 
-  # Space Complexity: 
-  def add(key, value)
-    raise NotImplementedError
+  # Time Complexity: O(logn)
+  # Space Complexity: O(logn)
+  def add(key, value = nil)
+    unless root
+      @root = TreeNode.new(key, value)
+    else
+      add_helper(@root, key, value)
+    end
   end
 
-  # Time Complexity: 
-  # Space Complexity: 
+  def add_helper(current, key, value)
+    return TreeNode.new(key, value) unless current
+
+    if key < current.key
+      current.left = add_helper(current.left, key, value)
+    else
+      current.right = add_helper(current.right, key, value)
+    end
+
+    # still working through why this needed/what would happen with no return
+    return current
+    # i think needed bc you assign current.left/right
+    # to the return of add_helper
+    # this assignment is needed because you will need
+    # to connect the new node to a leaf node
+    # (set that leaf node's .left/.right)
+    # so for a given current node, you need to know what
+    # it should be linked to (ie it's .left/right)
+    # which means that on the next iteration, just return
+    # that current's value to fill this in later
+    # ^ bad explanation - but if current = 97, and current.left is 91
+    # you need to return 91 on the next recursive call (when current is 91)
+    # so that way when you work back up the callstack, you'll know that when
+    # current is 97, set 97.left to be 91 (bc 91 is returned from helper(97.left))
+
+    # if you didn't return current
+    # just return -> going back up the callstack would be setting current.left/right to nil
+    # no return -> i think bc of implicit returns in ruby, it would just carry the
+    # assignment from the base case all the way up
+    # so like in find for ex, once you find it, it just carries that val all the way up
+    # when you're assigning, it takes the old leaf nodes assignment of .left/.right and just carries
+    # that all the way up
+    # this is why when i was playing around with it (without a return), in the case of adding a 3rd(+) node,
+    # the code would just set what root.left/right's left/right to be the node i was adding
+    # it basically takes that assignment from the base case and carries it all the way up
+    # see ex on line 186, when you dont have a return after an if block, the method just returns what the if block returned (when you print it)
+
+  end
+
+  # Time Complexity: O(logn)
+  # Space Complexity: O(logn)
   def find(key)
-    raise NotImplementedError
+    return find_helper(@root, key)
+  end
+
+  def find_helper(current, key)
+    return current unless current
+    return current.value if key == current.key
+
+    if current.key < key
+      find_helper(current.right, key)
+    else
+      find_helper(current.left, key)
+    end
+
+    # Q: why isnt a return needed here?
+    # A: i think bc it just carries that return current all the way back up
+    # implicit returns in ruby
   end
 
-  # Time Complexity: 
-  # Space Complexity: 
+  # Time Complexity: O(n)
+  # Space Complexity: O(n)
   def inorder
-    raise NotImplementedError
+    tree_in_order = []
+    inorder_helper(tree_in_order, @root)
   end
 
-  # Time Complexity: 
-  # Space Complexity: 
+  def inorder_helper(tree, current)
+    return tree unless current
+
+    inorder_helper(tree, current.left)
+    tree.push({:key=>current.key, :value=>current.value})
+    inorder_helper(tree, current.right)
+
+    # return tree
+  end
+
+  # Time Complexity: O(n)
+  # Space Complexity: O(n)
   def preorder
-    raise NotImplementedError
+    tree_pre_order = []
+    return preorder_helper(@root, tree_pre_order)
+  end
+
+  def preorder_helper(current, tree)
+    return tree unless current
+
+    tree.push({:key=>current.key, :value=>current.value})
+    preorder_helper(current.left, tree)
+    preorder_helper(current.right, tree)
+
+    return tree
   end
 
-  # Time Complexity: 
-  # Space Complexity: 
+  # Time Complexity: O(n)
+  # Space Complexity: O(n)
   def postorder
-    raise NotImplementedError
+    tree_post_order = []
+    return postorder_helper(@root, tree_post_order)
   end
 
-  # Time Complexity: 
-  # Space Complexity: 
+  def postorder_helper(current, tree)
+    return tree unless current
+
+    postorder_helper(current.left, tree)
+    postorder_helper(current.right, tree)
+    tree.push({:key=>current.key, :value=>current.value})
+
+    return tree
+  end
+
+  # Time Complexity: O(n)
+  # Space Complexity: O(n)
   def height
-    raise NotImplementedError
+    return height_helper(@root)
+  end
+
+  def height_helper(current)
+    return 0 unless current
+
+    left_height = height_helper(current.left)
+    right_height = height_helper(current.right)
+
+    return left_height < right_height ? right_height + 1 : left_height + 1
   end
 
   # Optional Method
-  # Time Complexity: 
-  # Space Complexity: 
+  # Time Complexity: O(n*m)
+  # Space Complexity: O(n)
   def bfs
-    raise NotImplementedError
+    q = []
+    tree_bfs = []
+    return tree_bfs unless @root
+
+    q.push(@root)
+
+    until q.empty?
+      current = q.shift
+
+      tree_bfs.push({:key=>current.key, :value=>current.value})
+
+      q.push(current.left) if current.left
+      q.push(current.right) if current.right
+    end
+
+    return tree_bfs
+
   end
 
   # Useful for printing
   def to_s
     return "#{self.inorder}"
   end
 end
+
+
+def test(x, y)
+  if x == 1
+    x = 3
+    # p (y)
+  else
+    x = 4
+    # p ("no")
+  end
+  # return
+end
+
+# p test(1, 2)
+
+
+tree = Tree.new()
+tree.add(5, "5")
+tree.add(3, "3")
+tree.add(1, "1")
+#     5
+#   3
+# 1
+p tree.inorder
+
+tree = Tree.new()
+tree.add(1, "1")
+tree.add(3, "3")
+tree.add(5, "5")
+    #   1
+    # 3   5
+p tree.inorder

test/tree_test.rb

@@ -37,7 +37,6 @@
     end
 
     it "will return the tree in order" do
-
       expect(tree_with_nodes.inorder).must_equal [{:key=>1, :value=>"Mary"}, {:key=>3, :value=>"Paul"}, 
                                        {:key=>5, :value=>"Peter"}, {:key=>10, :value=>"Karla"}, 
                                        {:key=>15, :value=>"Ada"}, {:key=>25, :value=>"Kari"}]
@@ -95,8 +94,8 @@
       expect(my_tree.height).must_equal 1
     end
 
-    it "will report the height for a balanced tree" do
-      expect(tree_with_nodes.height).must_equal 3
+    it "will report the height for a tree with height 4 on right and 3 on left" do
+      expect(tree_with_nodes.height).must_equal 4
     end
 
     it "will report the height for unbalanced trees" do