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

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

本来想先对inorder array做预处理存上val对应的index，结果发现 val可能有duplicates。

 1 /**duplicates
 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[] preorder, int[] inorder) {
12         // IMPORTANT: Please reset any member data you declared, as
13         // the same Solution instance will be reused for each test case.
14         return preorder_inorder(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
15     }
16     public TreeNode preorder_inorder(int[] pre, int ps, int pe, int[] in, int is, int ie){
17         if(ps > pe || is > ie) return null;
18         TreeNode root = new TreeNode(pre[ps]);
19         int ind = 0;
20         for(int i = is; i <= ie; i++)
21             if(in[i] == root.val){
22                 ind = i;
23                 break;
24         }
25         int len = ind - is;
26         root.left = preorder_inorder(pre, ps + 1, ps + len, in, is, ind - 1);
27         root.right = preorder_inorder(pre, ps + 1 + len, pe, in, ind + 1, ie);
28         return root;
29     }
30 }