题目：

给定两个整数数组，preorder 和 postorder ，其中 preorder 是一个具有 无重复 值的二叉树的前序遍历，postorder 是同一棵树的后序遍历，重构并返回二叉树。

如果存在多个答案，您可以返回其中 任何 一个。

示例 1：

输入：preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
输出：[1,2,3,4,5,6,7]
示例 2:

输入: preorder = [1], postorder = [1]
输出: [1]

提示：

1 <= preorder.length <= 30
1 <= preorder[i] <= preorder.length
preorder 中所有值都 不同
postorder.length == preorder.length
1 <= postorder[i] <= postorder.length
postorder 中所有值都 不同
保证 preorder 和 postorder 是同一棵二叉树的前序遍历和后序遍历

java代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
 class Solution {
    public TreeNode constructFromPrePost(int[] pre, int[] post) {
        int n = pre.length;
        if (n == 0) {
            return null;
        }
        TreeNode root = new TreeNode(pre[0]);
        if (n == 1) {
            return root;
        }

        int l = 0;
        for (int i = 0; i < n; i++) {
            if (post[i] == pre[1]) {
                l = i + 1;
                break;
            }
        }

        int[] leftPre = new int[l];
        int[] leftPost = new int[l];
        int[] rightPre = new int[n - l - 1];
        int[] rightPost = new int[n - l - 1];

        for (int i = 1; i <= l; i++) {
            leftPre[i - 1] = pre[i];
        }

        for (int i = 0; i < l; i++) {
            leftPost[i] = post[i];
        }

        for (int i = l + 1; i < n; i++) {
            rightPre[i - l - 1] = pre[i];
        }

        for (int i = l; i < n - 1; i++) {
            rightPost[i - l] = post[i];
        }


        root.left = constructFromPrePost(leftPre, leftPost);
        root.right = constructFromPrePost(rightPre, rightPost);
        return root;

    }
}