输入两棵二叉树A，B，判断B是不是A的子结构

问题描述：输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

package hgy.java.arithmetic;

import java.util.Arrays;
import org.junit.Test;

public class HasSubtree {
	public class TreeNode {
		int val = 0;
		TreeNode left = null;
		TreeNode right = null;
		public TreeNode(int val) {
			this.val = val;
		}
	}

	// 通过树的先序数组和中序数组构造一棵二叉树
	public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
		if (pre.length == 0)
			return null;
		TreeNode root = new TreeNode(pre[0]);
		for (int i = 0; i < in.length; i++) {
			if (in[i] == pre[0]) {
				// copyOfRange复制Arrays的[from,to)内容;
				root.left = reConstructBinaryTree(
						Arrays.copyOfRange(pre, 1, i + 1),
						Arrays.copyOfRange(in, 0, i));
				root.right = reConstructBinaryTree(
						Arrays.copyOfRange(pre, i + 1, pre.length),
						Arrays.copyOfRange(in, i + 1, in.length));
				break;
			}
		}
		return root;
	}
	
	public boolean hasSubtree(TreeNode root1, TreeNode root2) {
		if (root1 == null || root2 == null)
			return false;
		if (root1.val == root2.val) {
			if (judge(root1, root2))
				return true;
		}
		// 判断左右子树是否包含root2
		return hasSubtree(root1.left, root2) || hasSubtree(root1.right, root2);
	}

	// 判断是否为子树
	public boolean judge(TreeNode root1, TreeNode root2) {
		if (root2 == null)
			return true;
		if (root1 == null)
			return false;
		if (root1.val == root2.val)
			return judge(root1.left, root2.left)
					&& judge(root1.right, root2.right);
		return false;
	}

	@Test
	public void test() {
		int[] pre = { 1, 2, 4, 7, 3, 5, 6, 8 };
		int[] in = { 4, 7, 2, 1, 5, 3, 8, 6 };
		TreeNode tree1 = reConstructBinaryTree(pre, in);
		int[] pre2 = { 1, 2, 4, 7 };
		int[] in2 = { 4, 7, 2, 1 };
		TreeNode tree2 = reConstructBinaryTree(pre2, in2);
		System.out.println(hasSubtree(tree1, tree2));
	}

}