leetcode------Construct Binary Tree from Inorder and Postorder Traversal

标题：	Construct Binary Tree from Inorder and Postorder Traversal
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：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder.length==0||postorder.length==0)return null;
        TreeNode root=new TreeNode(postorder[postorder.length-1]);
        int i=0;
        for(;i<inorder.length;i++){
            if(inorder[i]==postorder[postorder.length-1])break;
        }
        int [] new_pos_left,new_pos_right,new_in_left,new_in_right;
        if(i<postorder.length){
            new_in_left=new int[i];
            System.arraycopy(inorder, 0, new_in_left, 0, i); 
            new_pos_left=new int [i];
            System.arraycopy(postorder, 0, new_pos_left, 0, i); 
            root.left=buildTree(new_in_left,new_pos_left);
            
            new_in_right=new int [inorder.length-i-1];
            System.arraycopy(inorder, i+1, new_in_right, 0, inorder.length-i-1);
            new_pos_right=new int [postorder.length-i-1];
            System.arraycopy(postorder, i, new_pos_right, 0, postorder.length-i-1);
            root.right=buildTree(new_in_right,new_pos_right);
        }
        return root;
    }
}