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

标题：	Construct Binary Tree from Preorder and Inorder Traversal
通过率：	26.5
难度：	中等

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

/**
 * 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[] preorder, int[] inorder) {
        if(preorder.length==0||inorder.length==0){
            return null;
        }
        TreeNode root=new TreeNode(preorder[0]);
        int i=0;
        for(;i<inorder.length;i++){
            if(inorder[i]==preorder[0])break;
        }
        int [] new_pre_left,new_pre_right,new_in_left,new_in_right;
        if(i<preorder.length){
            new_in_left=new int[i];
            System.arraycopy(inorder, 0, new_in_left, 0, i); 
            new_pre_left=new int[i];
            System.arraycopy(preorder, 1, new_pre_left, 0, i);  
            root.left=buildTree(new_pre_left,new_in_left);
            
            new_in_right=new int[preorder.length-i-1];
            System.arraycopy(inorder, i+1, new_in_right, 0, preorder.length-i-1); 
            new_pre_right=new int[preorder.length-i-1];
             System.arraycopy(preorder, i+1, new_pre_right, 0, preorder.length-i-1); 
             root.right=buildTree(new_pre_right,new_in_right);
            
        }
        return root;
    }
}