树的子结构问题

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 oj;
/*思路：参考剑指offer
1、首先设置标志位result = false，因为一旦匹配成功result就设为true，
剩下的代码不会执行，如果匹配不成功，默认返回false
2、递归思想，如果根节点相同则递归调用DoesTree1HaveTree2（），
如果根节点不相同，则判断tree1的左子树和tree2是否相同，
再判断右子树和tree2是否相同
3、注意null的条件，HasSubTree中，如果两棵树都不为空才进行判断，
DoesTree1HasTree2中，如果Tree2为空，则说明第二棵树遍历完了，即匹配成功，
tree1为空有两种情况（1）如果tree1为空&&tree2不为空说明不匹配，
（2）如果tree1为空，tree2为空，说明匹配。
 
*/
class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

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

}

public class Solution {
    public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        boolean result = false;
        if(root1 != null && root2 != null){
            if(root1.val == root2.val){
                result = DoesTree1HaveTree2(root1,root2);}
                if(!result){result = HasSubtree(root1.left, root2);}
            if(!result){result = HasSubtree(root1.right, root2);}
        }
        return result;
	 }

    public boolean DoesTree1HaveTree2(TreeNode root1,TreeNode root2){
	     if(root1 == null && root2 != null) return false;
	    if(root2 == null) return true;
	 if(root1.val != root2.val) return false;
	    return DoesTree1HaveTree2(root1.left, root2.left) && DoesTree1HaveTree2(root1.right, root2.right);
	}
}

上述是按照剑指offer的思路去解这道题。这种题的另一种问法是这样的：

给定两个二叉树T1和T2，返回T1的某个子树结构是否与T2的结构相等。

某大神给出的解法是使用kmp算法+二叉树序列化的方式。有兴趣的同志可以看一下。我最近也在学习，可以互相交流一下。

public class Code_03_KMP_T1SubtreeEqualsT2 {

	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}

	public static boolean isSubtree(Node t1, Node t2) {
		String t1Str = serialByPre(t1);
		String t2Str = serialByPre(t2);
		return getIndexOf(t1Str, t2Str) != -1;
	}

	public static String serialByPre(Node head) {
		if (head == null) {
			return "#!";
		}
		String res = head.value + "!";
		res += serialByPre(head.left);
		res += serialByPre(head.right);
		return res;
	}

	// KMP
	public static int getIndexOf(String s, String m) {
		if (s == null || m == null || m.length() < 1 || s.length() < m.length()) {
			return -1;
		}
		char[] ss = s.toCharArray();
		char[] ms = m.toCharArray();
		int[] nextArr = getNextArray(ms);
		int index = 0;
		int mi = 0;
		while (index < ss.length && mi < ms.length) {
			if (ss[index] == ms[mi]) {
				index++;
				mi++;
			} else if (nextArr[mi] == -1) {
				index++;
			} else {
				mi = nextArr[mi];
			}
		}
		return mi == ms.length ? index - mi : -1;
	}

	public static int[] getNextArray(char[] ms) {
		if (ms.length == 1) {
			return new int[] { -1 };
		}
		int[] nextArr = new int[ms.length];
		nextArr[0] = -1;
		nextArr[1] = 0;
		int pos = 2;
		int cn = 0;
		while (pos < nextArr.length) {
			if (ms[pos - 1] == ms[cn]) {
				nextArr[pos++] = ++cn;
			} else if (cn > 0) {
				cn = nextArr[cn];
			} else {
				nextArr[pos++] = 0;
			}
		}
		return nextArr;
	}

	public static void main(String[] args) {
		Node t1 = new Node(1);
		t1.left = new Node(2);
		t1.right = new Node(3);
		t1.left.left = new Node(4);
		t1.left.right = new Node(5);
		t1.right.left = new Node(6);
		t1.right.right = new Node(7);
		t1.left.left.right = new Node(8);
		t1.left.right.left = new Node(9);

		Node t2 = new Node(2);
		t2.left = new Node(4);
		t2.left.right = new Node(8);
		t2.right = new Node(5);
		t2.right.left = new Node(9);

		System.out.println(isSubtree(t1, t2));

	}

}