completed Trees-2

constructTree.py

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+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : elements to the left and right of root in inorder array are used to form the left and right subtree.
+
+# 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 buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
+        def tree(left, right):
+            if left > right:
+                return None
+
+            root_val = postorder.pop()
+            root = TreeNode(root_val)
+
+            root.right = tree(inorder_idx_map[root_val] + 1, right)
+            root.left = tree(left, inorder_idx_map[root_val] - 1)
+
+            return root
+
+        inorder_idx_map = {}
+
+        for idx, val in enumerate(inorder):
+            inorder_idx_map[val] = idx
+
+        return tree(0, len(postorder) - 1)

sumRootToLeafNumbers.py

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+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : On reaching leaf node, return the number formed so far as it is a complete path.
+# Return the sum of results from left and right subtrees to get the final answer.
+
+# 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 helper(self, root: Optional[TreeNode], currNum: int) -> int:
+        if root == None:
+            return 0
+
+        currNum = currNum * 10 + root.val
+
+        if root.left is None and root.right is None:
+            return currNum
+
+        left = self.helper(root.left, currNum)
+        right = self.helper(root.right, currNum)
+
+        return left + right
+
+
+    def sumNumbers(self, root: Optional[TreeNode]) -> int:
+        return self.helper(root, 0)
+
+