t3

pathsum2.py

@@ -0,0 +1,29 @@
+# O(h) time, O(h) space
+# 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 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,path):
+        if root is None:
+            return
+
+        currSum += root.val
+        path.append(root.val)
+
+        if root.left is None and root.right is None:
+            if currSum == targetSum:
+                # We append a copy of the path using list() 
+                # so it doesn't change when we pop later
+                self.result.append(list(path))
+
+        self.helper(root.left,targetSum,currSum,path)
+        self.helper(root.right, targetSum, currSum,path)
+        path.pop()

symmetric-tree.py

@@ -0,0 +1,21 @@
+# O(n) time bc we visit every node
+# O(h) space bc max depth of recursion stack
+
+# compare the left child of each node to the right child on the other side of the tree
+
+# 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 isSymmetric(self, root: Optional[TreeNode]) -> bool:
+        if root is None:
+            return True 
+        return self.ismirror(root.left,root.right)
+
+    def ismirror(self, leftroot, rightroot):
+        if leftroot and rightroot:
+            return leftroot.val == rightroot.val and self.ismirror(leftroot.left, rightroot.right) and self.ismirror(leftroot.right,rightroot.left)
+        return leftroot == rightroot