Create 94_binary_tree_inorder_traversal.md

Author: Joseph Luce

leetcode/medium/94_binary_tree_inorder_traversal.md

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+# 94. Binary Tree Inorder Traversal
+
+## Recursive solution
+- Runtime: O(N)
+- Space: O(1)
+- N = Number of elements in tree
+
+Make sure you understand the recursive solution first before attempting the iterative one.
+The main point is to understand the recursive call and how it backtracks to the previous call when it reaches the base case.
+
+```
+class Solution:
+    def inorderTraversal(self, root: TreeNode) -> List[int]:
+        def inorder_traversal_helper(root, result):
+            if root is None:
+                return
+            inorder_traversal_helper(root.left, result)
+            result.append(root.val)
+            inorder_traversal_helper(root.right, result)
+            
+        result = list()
+        inorder_traversal_helper(root, result)
+        return result
+```
+
+## Iterative solution
+- Runtime: O(N)
+- Space: O(1)
+- N = Number of elements in tree
+
+To fully understand this implementation, I recommend you draw this out step by step.
+Iterative inorder traversal can be confusing, no amount of words can help describe its inner workings.
+Don't try to memorize this, understand it.
+
+A good starting point for converting a recursive binary tree solution into an iterative one is to start with a stack, current node variable and a while loop.
+Those should be your starting ingredients.
+
+```
+class Solution:
+    def inorderTraversal(self, root: TreeNode) -> List[int]:
+        stack, result = list(), list()
+        curr_node = root
+        while True:
+            if curr_node is not None:
+                stack.append(curr_node)
+                curr_node = curr_node.left
+            elif len(stack) > 0:
+                curr_node = stack.pop()
+                result.append(curr_node.val)
+                curr_node = curr_node.right
+            else:
+                break
+        return result
+```