Done Tree-3

Problem1.py

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+#Problem  113. PATH SUM II
+# TIME COMPELXITY: O(H) where H denotes the height of tree
+# SPACE COMPLEXITY: O(H) where H denotes the height of tree as we are making recursion stack based on the height of tree (since doing DFS) 
+
+# 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 pathSum(self, root, targetSum):
+        """
+        :type root: Optional[TreeNode]
+        :type targetSum: int
+        :rtype: List[List[int]]
+        """
+        self.result=[]
+        self.helper(root,targetSum,0,[])
+        return self.result
+
+    def helper(self,root,targetSum,currentSum,path):
+        #base
+        if root is None:
+            return
+
+        #logic
+        path.append(root.val)
+        currentSum+=root.val
+
+        if root.left is None and root.right is None and currentSum==targetSum:
+            self.result.append(list(path))
+
+
+        self.helper(root.left,targetSum,currentSum,path)
+        self.helper(root.right,targetSum,currentSum,path)
+        path.pop()

Problem2.py

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+#Problem 101. SYMMETRIC TREE
+# TIME COMPELXITY: O(N) as we need to compare each node N
+# SPACE COMPLEXITY: O(H) where H denotes the height of the tree as we are doing recursion stack
+
+# 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 isSymmetric(self, root):
+        """
+        :type root: Optional[TreeNode]
+        :rtype: bool
+        """
+        if root is None:
+            return True
+
+        return self.symmetricCheck(root.left,root.right)
+
+    def symmetricCheck(self,left,right):
+        if left is None and right is None:
+            return True
+
+        if left is None or right is None:
+            return False
+
+        if left.val!=right.val:
+            return False
+
+        return self.symmetricCheck(left.left,right.right) and self.symmetricCheck(left.right,right.left)
+