Update 617_merge_two_binary_trees.md

Author: Joseph Luce

leetcode/easy/617_merge_two_binary_trees.md

@@ -1,16 +1,18 @@
 # 617. Merge Two Binary Trees
 
-## Recursion Solution
+## Recursive Solution
 
 - Runtime: O(N)
-- Space: O(1) (Assuming creating the result is not extra space)
+- Space: O(N)
 - N = Number of elements in both trees
 
 By traversing both trees together, the solution is much simpler. 
 Use one of the trees as your primary merge tree during the recursion and create new nodes when the primary doesn't have one and the secondary tree does.
 With recursion, you need to create the new nodes when you are the parent before you perform a function call into that node.
 You cannot create the new node if the node doesn't exist after the call.
 
+Recursion always has O(N) at the very least.
+
 ```
 class Solution:
     def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
@@ -30,3 +32,32 @@ class Solution:
         merge_tree_helper(t1, t2)
         return t1
 ```
+
+## Iterative Solution
+
+- Runtime: O(N)
+- Space: O(N)
+- N = Number of elements in both trees
+
+Similar to the recursion solution, however, will need to keep two items in the each element of the stack since the idea was to traverse the nodes in pairs. You could use two stacks but I believe the code would be more clunky.
+
+```
+class Solution:
+    def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
+        if t1 is None:
+            return t2
+        stack = list()
+        stack.append((t1, t2))
+        while len(stack) > 0:
+            node1, node2 = stack.pop()
+            if node1 is None or node2 is None:
+                continue
+            node1.val += node2.val
+            if node2.left is not None and node1.left is None:
+                node1.left = TreeNode(0)
+            if node2.right is not None and node1.right is None:
+                node1.right = TreeNode(0)
+            stack.append((node1.left, node2.left))
+            stack.append((node1.right, node2.right))
+        return t1
+```