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);}}