思路：

        先统计B数的非空节点数countB。然后前序遍历A，当遇到A的值和B的第一个值相等时，则进行统计左右结构和值都相等的节点数和sum，如果sum == countB，则true。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countB = 0;
    public boolean flag = false;
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if(B == null || A == null) {
            return false;
        }
        count(B);
        identify(A, B);
        return flag;
    }

    public void identify(TreeNode A, TreeNode B) {
        if (flag || A == null || B == null) {
            return;
        }
        if (A.val == B.val ) {
            // 统计结构和值都相等的节点数
            int sum = countAB(A.left, B.left) + countAB(A.right, B.right) + 1;
            if(sum == countB) {
                flag = true;
                return;
            }
            
        } 
        identify(A.left, B);
        identify(A.right, B);
        return;
    } 

    public int countAB(TreeNode A, TreeNode B) {
        if (A == null || B == null) {
            return 0;
        }
        if(A.val == B.val) {
            return countAB(A.left, B.left) + countAB(A.right, B.right) + 1;
        }
        return 0;
    }

    public void count(TreeNode t) {
        if (t == null) {
            return;
        }
        countB ++;
        count(t.left);
        count(t.right);
    }

    
}