Leetcode-Construct Binary Tree from inorder and preorder travesal

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:

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public TreeNode buildTree(int[] preorder, int[] inorder) {
         TreeNode root = buildTreeRecur(preorder,inorder,0,preorder.length-1,0,inorder.length-1);
         return root;
     }

     public TreeNode buildTreeRecur(int[] preorder, int[] inorder, int preS, int preE, int inS, int inE){
         if (preS>preE) return null;
         if (preS==preE){
             TreeNode leaf = new TreeNode(preorder[preS]);
             return leaf;
         }

         int rootVal = preorder[preS];
         int index = inS;
         for (int i=inS;i<=inE;i++)
             if (inorder[i]==rootVal){
                 index = i;
                 break;
             }

         int leftLen = index-inS;
         int rightLen = inE - index;

         TreeNode leftChild = buildTreeRecur(preorder,inorder,preS+1,preS+leftLen,inS,index-1);
         TreeNode rightChild = buildTreeRecur(preorder,inorder,preE-rightLen+1,preE,index+1,inE);
         TreeNode root = new TreeNode(rootVal);
         root.left = leftChild;
         root.right= rightChild;
         return root;
    }

 }

Construct Binary Tree from Preorder and Inorder Traversal