From 51685c1ba07a9f08f4a48b1980dc14f2ef20baa4 Mon Sep 17 00:00:00 2001
From: sreevardhan1099 <sreevardhan1099@gmail.com>
Date: Mon, 26 Jan 2026 00:46:23 -0500
Subject: [PATCH] Completed Trees-2

---
 Problem-1.py | 40 ++++++++++++++++++++++++++++++++++++++++
 Problem-2.py | 30 ++++++++++++++++++++++++++++++
 2 files changed, 70 insertions(+)
 create mode 100644 Problem-1.py
 create mode 100644 Problem-2.py

diff --git a/Problem-1.py b/Problem-1.py
new file mode 100644
index 00000000..aa6b210f
--- /dev/null
+++ b/Problem-1.py
@@ -0,0 +1,40 @@
+#106. Construct Binary Tree from Inorder and Postorder Traversal
+# Time Complexity :O(n)
+# Space Complexity :O(h)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+
+# Your code here along with comments explaining your approach
+#from postorder identify the root element from last element in aray
+# fint the root element index in the inorder array
+# Inorder array, everything on left of root is left subtree of root and everything on right is right subtree of root
+#make recusive calls to implement the above logic
+#time - O(n)
+#space - O(h)
+
+# 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]:
+        self.idx=len(postorder)-1 # index of postorder list
+        d=dict()##map for maintaining index of values in inorder
+        for i in range(0, len(inorder)):
+            d[inorder[i]]=i
+        return self.helper(postorder, d, 0,len(inorder)-1)
+    
+    def helper(self,postorder, d, start, end):
+        if(start>end):
+            return None
+
+        curr=postorder[self.idx]
+        print(self.idx)
+        root=TreeNode(curr)
+        self.idx-=1
+        root.right=self.helper(postorder, d, d[curr]+1, end)#right subtree of curr node
+        root.left=self.helper(postorder, d, start, d[curr]-1)#left subtree of curr node
+        return root
diff --git a/Problem-2.py b/Problem-2.py
new file mode 100644
index 00000000..cb911fa1
--- /dev/null
+++ b/Problem-2.py
@@ -0,0 +1,30 @@
+#129. Sum Root to Leaf Numbers
+# Time Complexity :O(n)
+# Space Complexity :O(h)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+
+# Your code here along with comments explaining your approach
+#resursively call the nodes in tree either is preorder,inorder,postorder. Maintin the sum at each node by passing sum as paramerter in recursion
+# 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 sumNumbers(self, root: Optional[TreeNode]) -> int:
+        self.res=0 # results as global variable
+        self.helper(root,0)#call helper passing initial sum as 0
+        return self.res
+    
+    def helper(self, root, sum):
+        #condition
+        if root is None:
+            return
+        sum=sum*10+root.val
+        if root.left is None and root.right is None: # if leaf node then add the sum to result
+            self.res+=sum
+        self.helper(root.left, sum)
+        self.helper(root.right, sum)