Done Tree-2

Problem1.py

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+#Problem 106. CONSTRUCT BINARY TREE FROM INORDER AND POSTORDER TRAVERSAL
+# TIME COMPELXITY: O(N) where N denotes the number of nodes ie the number of elements in a given array
+# SPACE COMPLEXITY: O(N) where N denotes the number of elements that is being recursively being called to make a stack call
+
+
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution(object):
+    def buildTree(self, inorder, postorder):
+        """
+        :type inorder: List[int]
+        :type postorder: List[int]
+        :rtype: Optional[TreeNode]
+        """
+        self.index=len(postorder)-1
+        inordermap={val:i for i,val in enumerate(inorder)}
+        return self.helper(postorder,0,len(postorder)-1,inordermap)
+
+    def helper(self,postorder,start,end,inordermap):
+        if start>end:
+            return None
+
+        root=postorder[self.index]
+        self.index=self.index-1
+        rootIndex=inordermap[root]
+
+        node=TreeNode(root)
+        node.right=self.helper(postorder,rootIndex+1,end,inordermap)
+        node.left=self.helper(postorder,start,rootIndex-1,inordermap)
+
+        return node

Problem2.py

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+#Problem 129. SUM ROOT TO LEAF NUMBERS
+# TIME COMPELXITY: O(N) where N denotes the number of nodes in a given tree
+# SPACE COMPLEXITY: O(H) where H denotes the height of the given tree as we are making a recursive stack call here
+
+
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution(object):
+    def sumNumbers(self, root):
+        """
+        :type root: Optional[TreeNode]
+        :rtype: int
+        """
+        #fullAns=0
+        return self.helper(root,0)
+
+    def helper(self,root,current):
+        #base
+        if root is None:
+            return 0
+
+        #logic
+        current=10*current+root.val
+
+        if root.left is None and root.right is None:
+            return current
+
+        left=self.helper(root.left,current)
+        right=self.helper(root.right,current)
+        #finalAns=left+right
+        return left+right
+