leetcode_105_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.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int []ArrPreorder;
	int []ArrInorder;
    Map<Integer, Integer>mapInorder=new HashMap<Integer, Integer>();
	public TreeNode buildTree(int[] preorder, int[] inorder) {
		if(preorder.length==0||preorder==null)
			return null;
		ArrPreorder=preorder;
    	ArrInorder=inorder;
    	for(int i=0;i<inorder.length;i++)
    		mapInorder.put(inorder[i], i);
        int start=0, end=preorder.length-1;
        TreeNode root =new TreeNode(0);
        createTree(root,start,end,start, end);
        return root;
    }
	public void createTree(TreeNode root, int start1,int end1,int start2,int end2)
	{
		int subStart1=start1,subStart2=start2,subEnd1=end1,subEnd2=end2;
		int target=ArrPreorder[start1];
		int indexInOrder=mapInorder.get(target);
		int len=indexInOrder-start2;
		int indexPreOrder=start1+len;
		if(start2<=indexInOrder-1)
		{
			subStart1=start1+1;
			subEnd1=indexPreOrder;
			subEnd2=indexInOrder-1;
			root.left=new TreeNode(0);
			createTree(root.left,subStart1,start2, subStart2, subEnd2);
		}
		root.val=target;
		if(indexInOrder+1<=end2)
		{
			subStart1=indexPreOrder+1;
			subStart2=indexInOrder+1;
			root.right=new TreeNode(0);
			createTree(root.right,subStart1,end1, subStart2, end2);
		}
	}
}