Commit a3746bc

@@ -0,0 +1,99 @@
+<<-DOC
+Note: Your solution should have O(inorder.length) complexity, since this is what you will be asked to accomplish in an interview.
+
+Let's define inorder and preorder traversals of a binary tree as follows:
+
+Inorder traversal first visits the left subtree, then the root, then its right subtree;
+Preorder traversal first visits the root, then its left subtree, then its right subtree.
+For example, if tree looks like this:
+
+    1
+   / \
+  2   3
+ /   / \
+4   5   6
+then the traversals will be as follows:
+
+Inorder traversal: [4, 2, 1, 5, 3, 6]
+Preorder traversal: [1, 2, 4, 3, 5, 6]
+Given the inorder and preorder traversals of a binary tree t, but not t itself, restore t and return it.
+
+Example
+
+For inorder = [4, 2, 1, 5, 3, 6] and preorder = [1, 2, 4, 3, 5, 6], the output should be
+restoreBinaryTree(inorder, preorder) = {
+    "value": 1,
+    "left": {
+        "value": 2,
+        "left": {
+            "value": 4,
+            "left": null,
+            "right": null
+        },
+        "right": null
+    },
+    "right": {
+        "value": 3,
+        "left": {
+            "value": 5,
+            "left": null,
+            "right": null
+        },
+        "right": {
+            "value": 6,
+            "left": null,
+            "right": null
+        }
+    }
+}
+For inorder = [2, 5] and preorder = [5, 2], the output should be
+restoreBinaryTree(inorder, preorder) = {
+    "value": 5,
+    "left": {
+        "value": 2,
+        "left": null,
+        "right": null
+    },
+    "right": null
+}
+Input/Output
+
+[time limit] 4000ms (rb)
+[input] array.integer inorder
+
+An inorder traversal of the tree. It is guaranteed that all numbers in the tree are pairwise distinct.
+
+Guaranteed constraints:
+1 ≤ inorder.length ≤ 2 · 103,
+-105 ≤ inorder[i] ≤ 105.
+
+[input] array.integer preorder
+
+A preorder traversal of the tree.
+
+Guaranteed constraints:
+preorder.length = inorder.length,
+-105 ≤ preorder[i] ≤ 105.
+
+[output] tree.integer
+
+The restored binary tree.
+DOC
+
+describe "#restore_binary_tree" do
+  def restore_binary_tree(inorder, preorder)
+    {}
+  end
+
+  null = nil
+  [
+    { inorder: [4, 2, 1, 5, 3, 6], preorder: [1, 2, 4, 3, 5, 6], x: { "value": 1, "left": { "value": 2, "left": { "value": 4, "left": null, "right": null }, "right": null }, "right": { "value": 3, "left": { "value": 5, "left": null, "right": null }, "right": { "value": 6, "left": null, "right": null } } } },
+    { inorder: [2, 5], preorder: [5, 2], x: { "value": 5, "left": { "value": 2, "left": null, "right": null }, "right": null } },
+    { inorder: [100], preorder: [100], x: { "value": 100, "left": null, "right": null } },
+    { inorder: [-100000, -99999, -99998], preorder: [-99999, -100000, -99998], x: { "value": -99999, "left": { "value": -100000, "left": null, "right": null }, "right": { "value": -99998, "left": null, "right": null } } },
+  ].each do |x|
+    it do
+      expect(restore_binary_tree(x[:inorder], x[:preorder])).to eql(x[:x])
+    end
+  end
+end