剑指offer17：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）

1 题目描述

　　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）

2 思路和方法

（1）先在A中找和B的根节点相同的结点

（2）找到之后遍历对应位置的其他结点，直到B中结点遍历完，都相同时，则B是A的子树

（3）对应位置的结点不相同时，退出继续在A中寻找和B的根节点相同的结点，重复步骤，直到有任何一棵二叉树为空退出

3 C++核心代码

3.1 递归实现1

 /*
 struct TreeNode {
     int val;
     struct TreeNode *left;
     struct TreeNode *right;
     TreeNode(int x) :
             val(x), left(NULL), right(NULL) {
     }
 };*/
 class Solution {
 public:
     bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
     {
         //1.先找到和子树根节点相同的结点
         bool flag=false;
         if((pRoot1!=NULL)&&(pRoot2!=NULL))
         {
             if(pRoot1->val==pRoot2->val)
                 //开始判断，此时找到了和子树根节点相同的结点
                 flag=HasSubtreetty(pRoot1, pRoot2);

             if(!flag)

                 flag=HasSubtree(pRoot1->left, pRoot2);

             if(!flag)

                 flag=HasSubtree(pRoot1->right, pRoot2);
         }
         return flag;
     }

     bool HasSubtreetty(TreeNode* pRoot1, TreeNode* pRoot2)
     {
         //该函数需要判断在找到和子树根节点相同的结点之后，判断其余结点是否相同
         if(pRoot2==NULL)
             return true;
         if((pRoot1==NULL)&&(pRoot2!=NULL))
             return false;
         if(pRoot1->val!=pRoot2->val)
             return false;
         return HasSubtreetty(pRoot1->left, pRoot2->left)&&HasSubtreetty(pRoot1->right, pRoot2->right);
     }
 };

3.2 递归实现2

 /*
 struct TreeNode {
     int val;
     struct TreeNode *left;
     struct TreeNode *right;
     TreeNode(int x) :
             val(x), left(NULL), right(NULL) {
     }
 };*/
 class Solution {
 public:
     bool IsSubtree(TreeNode* p1, TreeNode* p2)
     {
         if(p2==NULL)return ;
         if(p1==NULL)return ;

         if(p1->val != p2->val)return ;

         return IsSubtree(p1->left,p2->left) && IsSubtree(p1->right,p2->right);
     }
     bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
     {
         if(pRoot1==NULL || pRoot2==NULL)return ;

         return IsSubtree(pRoot1,pRoot2)||
             HasSubtree(pRoot1->left,pRoot2) ||
             HasSubtree(pRoot1->right,pRoot2);
     }
 };

4 完整代码

 #include <iostream>

 using namespace std;

 //struct ListNode {
 //    int val;
 //    struct ListNode *next;
 //    ListNode(int x) : val(x), next(NULL) {}
 //};

 struct TreeNode {
     int val;
     struct TreeNode *left;
     struct TreeNode *right;
     TreeNode(int x) :
         val(x), left(NULL), right(NULL) {
     }
 };

 class Solution {
 public:
     bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
     {
         //1.先找到和子树根节点相同的结点
         bool flag = false;
         if ((pRoot1 != NULL) && (pRoot2 != NULL))
         {
             if (pRoot1->val == pRoot2->val)
                 //开始判断，此时找到了和子树根节点相同的结点
                 flag = HasSubtreetty(pRoot1, pRoot2);

             if (!flag)

                 flag = HasSubtree(pRoot1->left, pRoot2);

             if (!flag)

                 flag = HasSubtree(pRoot1->right, pRoot2);
         }
         return flag;
     }

     bool HasSubtreetty(TreeNode* pRoot1, TreeNode* pRoot2)
     {
         //该函数需要判断在找到和子树根节点相同的结点之后，判断其余结点是否相同
         if (pRoot2 == NULL)
             return true;
         if ((pRoot1 == NULL) && (pRoot2 != NULL))
             return false;
         if (pRoot1->val != pRoot2->val)
             return false;
         return HasSubtreetty(pRoot1->left, pRoot2->left) && HasSubtreetty(pRoot1->right, pRoot2->right);
     }
 };

 int main()
 {
     Solution *s = new Solution();
     TreeNode *t1 = new TreeNode();
     TreeNode *t2 = new TreeNode();
     TreeNode *t3 = new TreeNode();
     TreeNode *t4 = new TreeNode();
     TreeNode *t5 = new TreeNode();
     TreeNode *t6 = new TreeNode();
     TreeNode *t7 = new TreeNode();
     TreeNode *t8 = new TreeNode();
     TreeNode *t9 = new TreeNode();
     t1->left = t2; t1->right = t3;
     t2->left = t4; t2->right = t5; t3->left = t6; t3->right = t7;
     t4->left = t8; t4->right = t9;

     TreeNode *tt1 = new TreeNode();    //只有8 6相同时也是子树，返回值为1
     //TreeNode *tt2 = new TreeNode(6);
     //TreeNode *tt3 = new TreeNode(5);
     TreeNode *tt2 = new TreeNode();
     TreeNode *tt3 = new TreeNode();
     tt1->left = tt2; tt1->right == tt3;

     bool out_tree = s->HasSubtree(t1, tt1);
     cout << out_tree << endl;

     system("pause");
     return ;
 }

参考资料

https://blog.csdn.net/weixin_36125166/article/details/75939373

https://blog.csdn.net/danxibaoxxx/article/details/93402407