Create 106_construct_binary_tree_from_inorder_and_postorder_traversal.md

Author: Joseph Luce

leetcode/medium/106_construct_binary_tree_from_inorder_and_postorder_traversal.md

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+# 106. Construct Binary Tree from Inorder and Postorder Traversal
+
+## Recursive Solution
+- Runtime: O(N)
+- Space: O(N) (Due to hash table)
+- N = Number of elements in list
+
+Similar to question 105.
+Postorder allows us to know which is the root node by using the last element of the array.
+With this, we can figure out the left and right sub-trees in the inorder traversal.
+Using recursion, we can continue to break up the sub-trees once we know which is the root of the sub-tree using this method.
+
+To allow for quicker look ups for roots, we can build an enmuerated hash table to find the indexes for each value of the inorder traversal.
+
+```
+class Solution:
+    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
+        
+        def build_helper(inorder_start, inorder_end, postorder_start, postorder_end):
+            if inorder_start > inorder_end or postorder_start > postorder_end:
+                return None
+            inorder_root_idx = val_to_inorder_idx[postorder[postorder_end]]
+            right_size = inorder_end - inorder_root_idx
+            node = TreeNode(postorder[postorder_end])
+            node.left = build_helper(inorder_start, inorder_root_idx-1, postorder_start, postorder_end - right_size - 1)
+            node.right = build_helper(inorder_root_idx + 1, inorder_end, postorder_end - right_size, postorder_end - 1)
+            return node
+        
+        val_to_inorder_idx = {val: i for i, val in enumerate(inorder)}
+        return build_helper(0, len(inorder)-1, 0, len(postorder)-1)
+```