题目说明B树如果为空树则不是A树的子结构 

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

    public boolean search(TreeNode A, TreeNode B){
        if(A==null) return false;
//比较该节点的值是否和B根节点相同，如果相同再判断是子树不是相同结构
        if(A.val==B.val&&compare(A,B)){
            return true;
        }
//遍历左子树，遍历右子树
        return search(A.left,B)||search(A.right,B);
    }

    public boolean compare(TreeNode A, TreeNode B){
        if(B==null){
            return true;
        }
        if(A==null){
            return false;
        }
 //判断当前节点相不相等，左右子树相不相等
        return A.val==B.val&&compare(A.left,B.left)&&compare(A.right,B.right);
    }
}

优化之后代码如下：

class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if(A == null || B == null) return false;
        return compare(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B);
    }
    public boolean compare(TreeNode A, TreeNode B){
        if(B == null) return true;
        if(A == null) return false;
        return A.val == B.val && compare(A.left, B.left) && compare(A.right, B.right);
    }
}