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：

 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<inorder.length;i++){

16             if(inorder[i]==postorder[postorder.length-1])break;

17         }

18         int [] new_pos_left,new_pos_right,new_in_left,new_in_right;

19         if(i<postorder.length){

20             new_in_left=new int[i];

21             System.arraycopy(inorder, 0, new_in_left, 0, i); 

22             new_pos_left=new int [i];

23             System.arraycopy(postorder, 0, new_pos_left, 0, i); 

24             root.left=buildTree(new_in_left,new_pos_left);

25             

26             new_in_right=new int [inorder.length-i-1];

27             System.arraycopy(inorder, i+1, new_in_right, 0, inorder.length-i-1);

28             new_pos_right=new int [postorder.length-i-1];

29             System.arraycopy(postorder, i, new_pos_right, 0, postorder.length-i-1);

30             root.right=buildTree(new_in_right,new_pos_right);

31         }

32         return root;

33     }

34 }