剑指 Offer 26. 树的子结构

题目链接：
https://leetcode.cn/problems/shu-de-zi-jie-gou-lcof/description/?favorite=xb9nqhhg&orderBy=hot

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
int flag=0;
int ff=1;
    void check(TreeNode *A,TreeNode *B)
    {
        if(ff==0) return ;
         if((A->val!=B->val)|| (A->left==NULL && B->left!=NULL) || (A->right==NULL && B->right!=NULL))
        {
            ff=0;
            return ;
        }
        if(A->left!=NULL&&B->left!=NULL) 
        {
            check(A->left,B->left);
        }
        if(A->right!=NULL&&B->right!=NULL)
        {
            check(A->right,B->right);
        }
        if((A->right!=NULL&&B->right==NULL)||(A->left!=NULL&&B->left==NULL))
        {
            return ;
        }
    }
    void panduan(TreeNode *A,TreeNode *B)
    {
        if(A==NULL) return ;
        if(A->val==B->val)
        {
           printf("%d\n",A->val);
            ff=1;
            check(A,B);
            if(ff==1)
            {
                flag =1;
            }
        }
        if(A->right!=NULL) 
        {
             panduan(A->right,B);
        }
       
        if(A->left!=NULL) panduan(A->left,B);
    }
    bool isSubStructure(TreeNode* A, TreeNode* B) {
        if(B==NULL) return false;
        panduan(A,B);
        if(flag) return true;
        else return false;
    }
};