Completed Trees-3

PathSum2.java

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+// Time Complexity : O(n)
+// Space Complexity : O(H) , height of the tree
+// 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
+/*
+Create a helper function which would keep track of currentsum by adding root's value in each recursion
+and also keeping track of the visited values through a list and recurse the same process for both left and
+right children. We will only add the visited path list to our result on the case of leaf nodes and when
+target sum is met. We also need to traverse through right branch after we hit leaf node of a specific root,
+so, we need to backtrack by 1 position so we can compute that.
+ */
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    List<List<Integer>> result;
+    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
+        this.result = new ArrayList<>();
+        helper(root, targetSum, 0 , new ArrayList<>());
+        return result;
+    }
+
+    private void helper(TreeNode root, int targetSum, int currSum, List<Integer> path) {
+        if(root == null)
+            return;
+
+        //action
+        currSum += root.val;
+        path.add(root.val);
+
+        if(root.left == null && root.right == null) {
+            if(currSum == targetSum) {
+                result.add(new ArrayList<>(path));
+            }
+        }
+
+        //recurse
+        helper(root.left, targetSum, currSum, path);
+        helper(root.right, targetSum, currSum, path);
+
+        //backtrack
+        path.remove(path.size() - 1);
+
+    }
+}

SymmetricTree.java

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+// Time Complexity : O(n)
+// Space Complexity : O(n)
+// 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
+/*
+Have a recursive function where we try to compare both the left and right branches of the tree along with
+their values and also the right and left branches as well to check if they are mirror images to each other
+or not. If we are able to traverse both of them until leaf nodes at the same time, it means both are of
+same height, if not, we return false.
+ */
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    public boolean isSymmetric(TreeNode root) {
+        return isMirror(root, root);
+    }
+
+    private boolean isMirror(TreeNode t1, TreeNode t2) {
+        if(t1 == null && t2 == null)
+            return true;
+        if(t1 == null || t2 == null)
+            return false;
+        return ((t1.val == t2.val) && isMirror(t1.left, t2.right) && isMirror(t1.right, t2.left));
+
+    }
+}