refactor 106

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_106.java

@@ -5,22 +5,12 @@
 import java.util.HashMap;
 import java.util.Map;
 
-/**
- * 106. Construct Binary Tree from Inorder and Postorder Traversal
- *
- * Given inorder and postorder traversal of a tree, construct the binary tree.
-
- Note:
- You may assume that duplicates do not exist in the tree.
- */
 public class _106 {
     public static class Solution1 {
 
         /**
          * https://discuss.leetcode.com/topic/3296/my-recursive-java-code-with-o-n-time-and-o-n-space
-         *
          * Note: the last element of postorder array is the root!
-         *
          * The idea is to take the last element in postorder as the root; find the position of the root
          * in the inorder array; then locate the range for left sub-tree and right sub-tree and do
          * recursion, use a hashmap to record the index of root in the inorder array.
@@ -35,11 +25,11 @@ public TreeNode buildTree(int[] inorder, int[] postorder) {
             }
             /**At the beginning, both start from 0 to nums.length-1*/
             return buildTreeRecursively(inorderMap, 0, inorder.length - 1, postorder, 0,
-                postorder.length - 1);
+                    postorder.length - 1);
         }
 
         private TreeNode buildTreeRecursively(Map<Integer, Integer> inorderMap, int inorderStart,
-            int inorderEnd, int[] postorder, int postorderStart, int postorderEnd) {
+                                              int inorderEnd, int[] postorder, int postorderStart, int postorderEnd) {
             if (postorderStart > postorderEnd || inorderStart > inorderEnd) {
                 return null;
             }