题目描述：

给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

示例 1:

输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]

示例 2:

输入：inorder = [-1], postorder = [-1]
输出：[-1]

代码（递归）：

/**
 * 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 buildTree(int[] inorder, int[] postorder) {
        if(inorder.length==0){
            return null;
        }
        int myValue=postorder[postorder.length-1];
        TreeNode parent=new TreeNode(myValue);
        for(int i=0;i<inorder.length;i++){
            if(inorder[i]==myValue){
                int[] inleftChild = Arrays.copyOfRange(inorder, 0, i);
                int[] inrightChild = Arrays.copyOfRange(inorder, i + 1, inorder.length);

                int[] postRight = Arrays.copyOfRange(postorder, i, postorder.length - 1);
                int[] postLeft = Arrays.copyOfRange(postorder, 0, i);

                parent.left=buildTree(inleftChild,postLeft);
                parent.right=buildTree(inrightChild,postRight);
                break;
            }
        }
        return parent;
    }
}