Added Day-27 Solution

Author: Gangadharbhuvan

Day-27_Construct_Binary_Tree_From_Inorder_and_Postorder_Traversal.py

@@ -0,0 +1,61 @@
+'''
+    Given inorder and postorder traversal of a tree, construct the binary tree.
+
+Note:
+You may assume that duplicates do not exist in the tree.
+
+For example, given
+
+inorder = [9,3,15,20,7]
+postorder = [9,15,7,20,3]
+Return the following binary tree:
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+
+
+
+'''
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def formTree(self, postorder, inorder, start, end):
+        if start > end:
+            return None
+        
+        root = TreeNode(postorder[self.postIndex])
+        self.postIndex -= 1 #decrement
+
+        if root is None:
+            return None
+          
+        if start == end:
+            return root
+        
+        index = self.inorder_hashmap[root.val]
+
+        #traversal
+        root.right = self.formTree(postorder, inorder, index+1, end)
+        root.left = self.formTree(postorder, inorder, start, index - 1)
+        return root
+    
+    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
+        #preorder -> Root, Left, Right
+        #inorder -> Left, root, Right
+        #postorder -> Left, Right, Root
+
+        self.inorder_hashmap = {}
+        self.postIndex = len(postorder) - 1
+
+        for num, i in enumerate(inorder):
+            self.inorder_hashmap[i] = num
+        
+        return self.formTree(postorder, inorder, 0, len(inorder)-1)