重建二叉树

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

思路：因为利用前序和后序来重建二叉树，前序是先根节点，后左子树，再右子树，中序是先左子树，后根节点，再右子树，所以前序的第一个一定是根节点，从中序里面找到根节点，左边的就是左子树，右边的就是右子树，不断地把数组切成一个个小的数组，直到为空。我的 思路是通过递归来实现

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

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

public class Solution1 {
    TreeNode t;

    public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
         if(pre.length == 0||in.length == 0){
            return null;
         }
            t = new TreeNode(pre[0]);

            for (int i = 0; i < in.length; i++) {
                if (in[i] == pre[0])

                {
                    t.left = reConstructBinaryTree(
                            Arrays.copyOfRange(pre, 1, i + 1),
                            Arrays.copyOfRange(in, 0, i));
                    t.right = reConstructBinaryTree(
                            Arrays.copyOfRange(pre, i + 1, pre.length),
                            Arrays.copyOfRange(in, i + 1, in.length));

                }
            }
        
        return t;
    }

但是我写的好像太占用空间，下面是借鉴了别人的写法

import java.util.Arrays;

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

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

public class Solution1 {

    public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
        return reConstructBinaryTree(pre, 0, pre.length - 1, in, 0,
                in.length - 1);

    }

    private TreeNode reConstructBinaryTree(int[] pre, int startPre, int endPre,
            int[] in, int startIn, int endIn) {

        if (startPre > endPre || startIn > endIn)
            return null;
        TreeNode root = new TreeNode(pre[startPre]);

        for (int i = startIn; i <= endIn; i++) {
            if (in[i] == pre[startPre]) {
                root.left = reConstructBinaryTree(pre, startPre + 1, startPre
                        + i - startIn, in, startIn, i - 1);
                root.right = reConstructBinaryTree(pre, i - startIn + startPre
                        + 1, endPre, in, i + 1, endIn);
            }
        }

        return root;
    }

    public static void main(String[] args) {
        int[] pre = { 1, 2, 4, 7, 3, 5, 6, 8 };
        int[] in = { 4, 7, 2, 1, 5, 3, 8, 6 };
        Solution1 s = new Solution1();
        s.reConstructBinaryTree(pre, in);
    }
}