【树】889. 根据前序和后序遍历构造二叉树

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public TreeNode constructFromPrePost(int[] pre, int[] post) 
12     {
13         int N = pre.length;
14         if (N == 0) 
15         {
16             return null;
17         }
18         TreeNode root = new TreeNode(pre[0]);
19         if (N == 1) 
20         {
21             return root;
22         }
23 
24         int L = 0;
25         for (int i = 0; i < N; ++i)
26             if (post[i] == pre[1])
27                 L = i+1;
28 
29         root.left = constructFromPrePost(Arrays.copyOfRange(pre, 1, L+1),
30                                          Arrays.copyOfRange(post, 0, L));
31         root.right = constructFromPrePost(Arrays.copyOfRange(pre, L+1, N),
32                                           Arrays.copyOfRange(post, L, N-1));
33         return root;
34 
35     }
36 }