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

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

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

根据前序遍历和中序遍历构建二叉树：看一棵树如下:

    1
   /   2   4
   \      5    6前序为：12546中序为：25146

从前序得知第一个数一定是树的root，通过中序得到右树和左树

递归调用即可，其中要注意递归的条件个字符串拷贝时的用法，

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[] preorder, int[] inorder) {
12         if(preorder.length==0||inorder.length==0){
13             return null;
14         }
15         TreeNode root=new TreeNode(preorder[0]);
16         int i=0;
17         for(;i<inorder.length;i++){
18             if(inorder[i]==preorder[0])break;
19         }
20         int [] new_pre_left,new_pre_right,new_in_left,new_in_right;
21         if(i<preorder.length){
22             new_in_left=new int[i];
23             System.arraycopy(inorder, 0, new_in_left, 0, i);
24             new_pre_left=new int[i];
25             System.arraycopy(preorder, 1, new_pre_left, 0, i);
26             root.left=buildTree(new_pre_left,new_in_left);
27
28             new_in_right=new int[preorder.length-i-1];
29             System.arraycopy(inorder, i+1, new_in_right, 0, preorder.length-i-1);
30             new_pre_right=new int[preorder.length-i-1];
31              System.arraycopy(preorder, i+1, new_pre_right, 0, preorder.length-i-1);
32              root.right=buildTree(new_pre_right,new_in_right);
33
34         }
35         return root;
36     }
37 }