标题：
通过率：	26.7%
难度：	中等

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

前面做过一个已经前序和中序求二叉树，本题是一直中序和后续求二叉树道理一样

前序的第一个值一定是树的root，那么后序的最后一个值一定是树的root，

具体看代码：

1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     public TreeNode buildTree(int[] inorder, int[] postorder) {12         if(inorder.length==0||postorder.length==0)return null;13         TreeNode root=new TreeNode(postorder[postorder.length-1]);14         int i=0;15         for(;i