Create 1008_construct_binary_search_tree_from_preorder_traversal.md

Author: Joseph Luce

leetcode/medium/1008_construct_binary_search_tree_from_preorder_traversal.md

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+## Recursive Solution
+
+- Runtime: O(N^2)
+- Space: O(1)
+- N = Number of elements in array
+
+I've selected this question because it would be good to know at least one reconstruction method of a BST.
+Preorder traversal was selected because it is one of the easier ones to understand.
+Unlike the other traversals, preorder has a property of knowing what the root node is by looking at the first element of the list.
+
+Since we are reconstructing a BST and not a binary tree, we can identify which sections are the left and right subtree by comparing their values to the first element.
+The left subtree will have values less/equal than the root and right subtree will have values greater than the root.
+
+```
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def bstFromPreorder(self, preorder: List[int]) -> TreeNode:
+        
+        def bst_helper(preorder):
+            if len(preorder) == 0:
+                return None
+            transition_index = len(preorder)
+            for index, node in enumerate(preorder):
+                if node > preorder[0]:
+                    transition_index = index
+                    break
+            root = TreeNode(preorder[0])
+            root.left = bst_helper(preorder[1:transition_index])
+            root.right = bst_helper(preorder[transition_index:])
+            return root
+            
+        return bst_helper(preorder)
+```