Trees-3

Problem1.py

@@ -0,0 +1,30 @@
+# 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
+
+# Time Complexity --> O(n)
+# Space Complexity --> O(log n)
+class Solution:
+    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
+        self.result = []
+        self.helper(root, targetSum, 0, [])
+        return self.result
+
+    def helper(self, root, targetSum, currSum, currList):
+        # base
+        if root is None:
+            return
+        # logic
+        currSum = currSum + root.val
+        currList.append(root.val)
+        if root.left is None and root.right is None:
+            if currSum==targetSum:
+                self.result.append(list(currList))
+
+        self.helper(root.left, targetSum, currSum, currList)
+        self.helper(root.right, targetSum, currSum, currList)
+        # backtrack
+        currList.pop()

Problem2.py

@@ -0,0 +1,25 @@
+# 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
+
+# Time Complexity --> O(n) where n is the number of nodes in the tree
+# Space Complexity --> O(log n)
+class Solution:
+    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
+        return self.helper(root.left, root.right)
+
+    def helper(self, n1, n2):
+        # base
+        if n1 is None and n2 is None:
+            return True
+        if (n1 is None and n2 is not None) or (n1 is not None and n2 is None):
+            return False
+        if n1.val!=n2.val:
+            return False
+        # logic
+        case1 = self.helper(n1.left, n2.right)
+        case2 = self.helper(n1.right, n2.left)
+        return case1 and case2