refactor 545

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_545.java

@@ -5,59 +5,6 @@
 import java.util.ArrayList;
 import java.util.List;
 
-/**
- * 545. Boundary of Binary Tree
- *
- * Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root.
- * Boundary includes left boundary, addLeaves, and right boundary in order without duplicate nodes.
- * Left boundary is defined as the path from root to the left-most node.
- * Right boundary is defined as the path from root to the right-most node.
- * If the root doesn't have left subtree or right subtree,
- * then the root itself is left boundary or right boundary.
- * Note this definition only applies to the input binary tree, and not applies to any subtrees.
- * The left-most node is defined as a leaf node you could reach when you always firstly travel
- * to the left subtree if exists. If not, travel to the right subtree.
- * Repeat until you reach a leaf node.
- * The right-most node is also defined by the same way with left and right exchanged.
-
- Example 1
- Input:
-  1
-   \
-   2
-  / \
- 3   4
-
- Ouput:
- [1, 3, 4, 2]
-
- Explanation:
- The root doesn't have left subtree, so the root itself is left boundary.
- The addLeaves are node 3 and 4.
- The right boundary are node 1,2,4. Note the anti-clockwise direction means you should output reversed right boundary.
- So order them in anti-clockwise without duplicates and we have [1,3,4,2].
-
-
- Example 2
- Input:
-    ____1_____
-   /          \
-  2           3
- / \          /
-4   5        6
-   / \      / \
-  7  8     9  10
-
- Ouput:
- [1,2,4,7,8,9,10,6,3]
-
- Explanation:
- The left boundary are node 1,2,4. (4 is the left-most node according to definition)
- The addLeaves are node 4,7,8,9,10.
- The right boundary are node 1,3,6,10. (10 is the right-most node).
- So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,6,3].
-
- */
 public class _545 {
     public static class Solution1 {
         public List<Integer> boundaryOfBinaryTree(TreeNode root) {