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.
Solution:
1. /*
2. * Definition for a binary tree node.
3. * public class TreeNode {
4. * int val;
5. * TreeNode left;
6. * TreeNode right;
7. * TreeNode(int x) { val = x; }
8. * }
9. /
10. public class Solution {
11. public TreeNode buildTree(int[] preorder, int[] inorder) {
12. Map map = new HashMap<>();
13. 14. for (int i = 0; i < inorder.length; i++) {
15. map.put(inorder[i], i);
16. }
17. 18. return build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, map);
19. }
20. 21. TreeNode build(int[] pre, int i1, int j1, int[] in, int i2, int j2, Map map) {
22. if (i1 > j1 || i2 > j2) {
23. return null;
24. }
25. 26. TreeNode root = new TreeNode(pre[i1]);
27. 28. int index = map.get(root.val);
29. int nleft = index - i2;
30. 31. root.left = build(pre, i1 + 1, i1 + nleft, in, i2, index - 1, map);
32. root.right = build(pre, i1 + nleft + 1, j1, in, index + 1, j2, map);
33. 34. return root;
35. }
36. }