update 1022.sum-of-root-to-leaf-binary-numbers.java

Author: ketankr9

1022.sum-of-root-to-leaf-binary-numbers.java

@@ -0,0 +1,73 @@
+/*
+ * @lc app=leetcode id=1022 lang=java
+ *
+ * [1022] Sum of Root To Leaf Binary Numbers
+ *
+ * https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/description/
+ *
+ * algorithms
+ * Easy (65.39%)
+ * Likes:    380
+ * Dislikes: 66
+ * Total Accepted:    33.7K
+ * Total Submissions: 51.6K
+ * Testcase Example:  '[1,0,1,0,1,0,1]'
+ *
+ * Given a binary tree, each node has value 0 or 1.  Each root-to-leaf path
+ * represents a binary number starting with the most significant bit.  For
+ * example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent
+ * 01101 in binary, which is 13.
+ * 
+ * For all leaves in the tree, consider the numbers represented by the path
+ * from the root to that leaf.
+ * 
+ * Return the sum of these numbers.
+ * 
+ * 
+ * 
+ * Example 1:
+ * 
+ * 
+ * 
+ * 
+ * Input: [1,0,1,0,1,0,1]
+ * Output: 22
+ * Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22
+ * 
+ * 
+ * 
+ * 
+ * Note:
+ * 
+ * 
+ * The number of nodes in the tree is between 1 and 1000.
+ * node.val is 0 or 1.
+ * The answer will not exceed 2^31 - 1.
+ * 
+ * 
+ */
+
+// @lc code=start
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+	int callme(int cur, TreeNode root){
+		if(root == null)
+			return 0;
+		cur = cur*2 + root.val;
+		if(root.left == null && root.right == null)
+			return cur;
+		return callme(cur, root.left) + callme(cur, root.right);
+	}
+    public int sumRootToLeaf(TreeNode root) {
+        return callme(0, root);
+    }
+}
+// @lc code=end