Problem : Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. We are using recursion to solve the problem .The tree is recursively called on the right subtree and then on the left subtree.A max level and current level counter is maintained.If the current level is gretaer than max level,we have reached the rightmost node of the level.

Java/binaryTreeRightSideView.java

@@ -0,0 +1,51 @@
+/**
+ * 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;
+ * }
+ * }
+ */
+/*
+Time Complexity : O(n)
+Space Complexity : O(n)
+
+Recursive call is made to right subtree and then the left subtree
+Note : if current level is greater than the max level we have visited rightmost node 
+*/
+
+class Solution {
+ static int maxLevel;
+ public static List<Integer> rightSideView(TreeNode root) {
+
+ maxLevel = 0;
+ int level = 1;
+ List<Integer> result = new ArrayList<Integer>();
+ rightSideViewHelper(root,level,result);
+
+ return result;
+ }
+ public static void rightSideViewHelper(TreeNode root,int level,List<Integer> result) 
+ {
+ if(root == null)
+ {
+ return; 
+ }
+
+ if(maxLevel < level)
+ {
+ maxLevel = level;
+ result.add(root.val);
+ }
+
+ rightSideViewHelper(root.right,level+1,result);
+ rightSideViewHelper(root.left,level+1,result);
+ }
+}

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@@ -490,7 +490,9 @@ DISCLAIMER: This above mentioned resources have affiliate links, which means if
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+ |
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