重建二叉树

题目描述：
根据二叉树的前序和中序，或者后序和中序，重建一颗二叉树
代码：

package offer;

class TreeNode{
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int val) {
        this.val = val;
    }
}


public class chongJianErChaShu{
    
    public static void main(String args[]) {
        
        int a[] = {1,2,4,7,3,5,6,8};
        int b[] = {4,7,2,1,5,3,8,6};
        int c[] = {7,4,2,5,8,6,3,1};
        TreeNode root = slove(a,0,a.length-1,b,0,b.length-1);
        TreeNode root2 = slove2(c,0,c.length-1, b,0,b.length-1);
        print(root);
        System.out.println();
        print(root2);
        
    }
    
    public static TreeNode slove(int a[],int astart, int aend, int b[], int bstart, int bend) {
        
        if (aend < astart || bend < bstart) {
            return null;
        }
        TreeNode root = new TreeNode(a[astart]);
        for (int i = 0; i<= bend; i++) {
            if (a[astart] == b[i]) {
                /* 参数传递这里有点绕人，绕了我好长时间，
                 * 可能自己真的脑筋转不过玩，最后把(bend-i)当做一个差值变量还好理解点。 */
                root.left = slove(a, astart+1, aend-(bend-i),b, bstart, i-1);
                root.right = slove(a,aend-(bend-i)+1 ,aend ,b, i+1,bend );
            }
        }
        return root;
    }
    public static TreeNode slove2(int a[], int astart, int aend, int b[], int bstart, int bend) {
        if (astart> aend || bstart > bend) {
            return null;
        }
        TreeNode root = new TreeNode(a[aend]);
        for (int i = 0; i<= bend; i++) {
            if (a[aend] == b[i]) {
                root.left = slove2(a, astart,aend-(bend-i)-1 ,b, bstart,i-1 );
                root.right = slove2(a,aend-(bend-i),aend-1,b, i+1,bend);
            }
            
            
        }
        return root;
    }    
    public static void print(TreeNode pHead) {
        if (pHead == null) {
            return ;
        }
        System.out.print(pHead.val+" ");
        print(pHead.left);
        print(pHead.right);
    }
    
    
}