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;}
}