refactor 105

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_105.java

@@ -5,51 +5,44 @@
 import java.util.HashMap;
 import java.util.Map;
 
-/**
- * 105. Construct Binary Tree from Preorder and Inorder Traversal
- * Given preorder and inorder traversal of a tree, construct the binary tree.
-
- Note:
- You may assume that duplicates do not exist in the tree.
- */
 public class _105 {
 
-	public static class Solution1 {
-		/**
-		 * credit: https://discuss.leetcode.com/topic/29838/5ms-java-clean-solution-with-caching use
-		 * HashMap as the cache so that accessing inorder index becomes O(1) time Note: The first
-		 * element of preorder array is the root!
-		 */
-		public TreeNode buildTree(int[] preorder, int[] inorder) {
-			Map<Integer, Integer> inorderMap = new HashMap();
-			for (int i = 0; i < inorder.length; i++) {
-				inorderMap.put(inorder[i], i);
-			}
-
-			/**At the beginning, both start from 0 to nums.length-1*/
-			return buildTree(preorder, 0, preorder.length - 1, inorderMap, 0, inorder.length - 1);
-		}
-
-		private TreeNode buildTree(int[] preorder, int preStart, int preEnd,
-				Map<Integer, Integer> inorderMap, int inStart, int inEnd) {
-			if (preStart > preEnd || inStart > inEnd) {
-				return null;
-			}
-
-			TreeNode root = new TreeNode(preorder[preStart]);
-			int inRoot = inorderMap.get(preorder[preStart]);
-			int numsLeft = inRoot - inStart;
-
-			/**It's easy to understand and remember:
-			 * for the indices of inorder array:
-			 * root.left should be inStart and inRoot-1 as new start and end indices
-			 * root.right should be inRoot+1 and inEnd as new start and end indices
-			 *
-			 * since inRoot is being used already in this recursion call, that's why we use inRoot-1 and inRoot+1
-			 * this part is the same for both Leetcode 105 and Leetcode 106.*/
-			root.left = buildTree(preorder, preStart + 1, preStart + numsLeft, inorderMap, inStart, inRoot - 1);
-			root.right = buildTree(preorder, preStart + numsLeft + 1, preEnd, inorderMap, inRoot + 1, inEnd);
-			return root;
-		}
-	}
+    public static class Solution1 {
+        /**
+         * credit: https://discuss.leetcode.com/topic/29838/5ms-java-clean-solution-with-caching use
+         * HashMap as the cache so that accessing inorder index becomes O(1) time Note: The first
+         * element of preorder array is the root!
+         */
+        public TreeNode buildTree(int[] preorder, int[] inorder) {
+            Map<Integer, Integer> inorderMap = new HashMap();
+            for (int i = 0; i < inorder.length; i++) {
+                inorderMap.put(inorder[i], i);
+            }
+
+            /**At the beginning, both start from 0 to nums.length-1*/
+            return buildTree(preorder, 0, preorder.length - 1, inorderMap, 0, inorder.length - 1);
+        }
+
+        private TreeNode buildTree(int[] preorder, int preStart, int preEnd,
+                                   Map<Integer, Integer> inorderMap, int inStart, int inEnd) {
+            if (preStart > preEnd || inStart > inEnd) {
+                return null;
+            }
+
+            TreeNode root = new TreeNode(preorder[preStart]);
+            int inRoot = inorderMap.get(preorder[preStart]);
+            int numsLeft = inRoot - inStart;
+
+            /**It's easy to understand and remember:
+             * for the indices of inorder array:
+             * root.left should be inStart and inRoot-1 as new start and end indices
+             * root.right should be inRoot+1 and inEnd as new start and end indices
+             *
+             * since inRoot is being used already in this recursion call, that's why we use inRoot-1 and inRoot+1
+             * this part is the same for both Leetcode 105 and Leetcode 106.*/
+            root.left = buildTree(preorder, preStart + 1, preStart + numsLeft, inorderMap, inStart, inRoot - 1);
+            root.right = buildTree(preorder, preStart + numsLeft + 1, preEnd, inorderMap, inRoot + 1, inEnd);
+            return root;
+        }
+    }
 }