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