How to reconstruct a tree(with no duplicate values) with pre-order and in-order sequencess of all nodes.

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode(int x) { val = x; }

* }

*/

public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder\) {

        if(preorder == null || inorder == null || preorder.length == 0 || inorder.length == 0 || preorder.length != inorder.length) {

            return null;

        }



        // use hashmap to store the inorder <value,index> pair

        Map<Integer,Integer> map = new HashMap<Integer,Integer>();

        for (int i = 0; i <inorder.length; i++) {

            map.put(inorder[i], i);

        }


        return helper(inorder, 0, inorder.length - 1, preorder, 0, preorder.length - 1, map);

    }



    private TreeNode helper(int[] in, int inleft, int inright, int[] pre, int preleft, int preright, Map<Integer,Integer> map) {

        // base case

        if(inleft > inright) {

            return null;

        }


        TreeNode root = new TreeNode(pre[preleft]);

        int size = map.get(root.val) - inleft; 



        root.left = helper(in, inleft, inleft + size - 1 , pre, preleft + 1, preleft + size, map);

        root.right = helper(in, inleft + size + 1, inright , pre, preleft + size + 1, preright, map);



        return root;

    }
}