Complete Trees-3 assignment

PathSumTwo.java

@@ -0,0 +1,35 @@
+//Time Complexity - O(n)
+//Space Complexity - O(h)
+class Solution {
+    List<List<Integer>> result = new ArrayList<>();
+    public List<List<Integer>> pathSum(TreeNode root, int targetSum)
+    {
+        //Validate the root
+        if (root == null) return new  ArrayList<>();
+        return helper(root, 0, targetSum, new ArrayList<>());
+    }
+
+    public List<List<Integer>> helper(TreeNode root, int currSum, int targetSum, List<Integer> path)
+    {
+        //base case
+        if (root == null) return new ArrayList<>();
+
+        //logic
+        path.add(root.val);
+        currSum += root.val;
+
+        if (root.left == null && root.right == null && currSum == targetSum)
+        {
+            result.add(new ArrayList(path));
+        }
+
+        //recurse
+        helper(root.left, currSum, targetSum, path);
+        helper(root.right, currSum, targetSum, path);
+
+        //back track
+        path.remove(path.size() - 1);
+
+        return result;
+    }
+}

SymetricTree.java

@@ -0,0 +1,24 @@
+//Approach: Using recursion we can check if the left value is not equal to right val, then return else or else the left and right subtrees are symmetric.
+//Time Complexity : O(n)
+//Space Complexity: O(1)
+class SymetricTree {
+    public boolean isSymmetric(TreeNode root)
+    {
+        if(root == null) return false;
+        return helper(root.left, root.right);
+    }
+
+    private boolean helper(TreeNode left, TreeNode right)
+    {
+        //base condition
+        if(left == null && right == null) return true;
+
+        //If any of the left or right nodes is null, that means the other part is not null and so not symmetric
+        if(left == null || right == null) return false;
+
+        //left value when equals to right value then it is considered symetric
+        if(left.val != right.val) return false;
+
+        return helper(left.left, right.right) && helper(left.right, right.left);
+    }
+}