Binary Trees2-Construct Tree Using Inorder and Preorder

For a given preorder and inorder traversal of a Binary Tree of type integer stored in an array/list, create the binary tree using the given two arrays/lists. You just need to construct the tree and return the root.
Note:
Assume that the Binary Tree contains only unique elements. 
Input Format:
The first line of input contains an integer N denoting the size of the list/array. It can also be said that N is the total number of nodes the binary tree would have.

The second line of input contains N integers, all separated by a single space. It represents the preorder-traversal of the binary tree.

The third line of input contains N integers, all separated by a single space. It represents the inorder-traversal of the binary tree.
Output Format:
The given input tree will be printed in a level order fashion where each level will be printed on a new line. 
Elements on every level will be printed in a linear fashion. A single space will separate them.
Constraints:
1 <= N <= 10^4
Where N is the total number of nodes in the binary tree.

Time Limit: 1 sec
Sample Input 1:
7
1 2 4 5 3 6 7 
4 2 5 1 6 3 7 
Sample Output 1:
1 
2 3 
4 5 6 7 
Sample Input 2:
6
5 6 2 3 9 10 
2 6 3 9 5 10 
Sample Output 2:
5 
6 10 
2 3 
9 

******************************************Code********************************************

public static BinaryTreeNode<Integer> buildTree(int[] preOrder, int[] inOrder) {
		
		BinaryTreeNode<Integer> root= buildTree(preOrder, inOrder,0,preOrder.length-1,0,inOrder.length-1);
		return root;
	}
  
	private static BinaryTreeNode<Integer> buildTree(int[] preOrder,int[] inOrder, int sPre, int ePre, int sIn, int eIn)
	{
		if(sPre>ePre)
			return null;
		int rootData = preOrder[sPre];
		BinaryTreeNode<Integer> root = new BinaryTreeNode<Integer>(rootData);

		int rootInorder = -1;
		for (int i = sIn; i <= eIn; i++) 
		{
			if (rootData == inOrder[i]) 
			{
				rootInorder = i;
				break;
			}
		}
		int sInLeft = sIn;
		int eInLeft = rootInorder - 1;

		int sPreLeft = sPre + 1;
		int leftSubTreeLength = eInLeft - sInLeft + 1;
		int ePreLeft = (sPreLeft) + (leftSubTreeLength - 1);

		int sInRight = rootInorder + 1;
		int eInRight = eIn;
		
		int sPreRight = ePreLeft + 1;
		int ePreRight = ePre;

		BinaryTreeNode<Integer> leftChild = buildTree(preOrder, inOrder, sPreLeft, ePreLeft, sInLeft, eInLeft);
		BinaryTreeNode<Integer> rightChild = buildTree(preOrder, inOrder, sPreRight, ePreRight, sInRight, eInRight);
		
		root.left = leftChild;
		root.right = rightChild;
		return root;
	}