How to tell if a binary tree tree is a subtree of another binary tree

#include <iostream>  #include <queue>  using namespace std;  struct treenode{  int value;  Treenode* left;  treenode* right;  treenode* parent;};/ /node structure void  Insert (treeNode **root,int value) {treenode* tobe=new treeNode;   tobe->value=value; tobe->parent=null; tobe->left=tobe->right=null; if (*root==null)  {   *root=tobe;       return;  }  treenode* Temp=*root,*p1=null;    while (temp)       {         p1=temp;         if (temp->value>value)           temp=temp->left;         else           temp=temp->right;       }    if (p1->value>value)       p1->left=tobe;    else       p1->right=tobe;    tobe->parent=p1;        }

Here is the relevant code <strong>bool Issametree (treenode* root1,treenode*root2) {if (root2==null&&root1==null)// Two equals NULL return true;if (root1==null| |    Root2==null)//Only one equals NULL return false; if (Root1->value==root2->value) return Issametree (Root1->left,root2->left) &&issametree (root1-    >right,root2->right); else return false;}    BOOL Issubtree (treenode* Root1,treenode *root2) {bool Issub=false;       if (root1!=null) {if (root1->value==root2->value) {issub=issametree (ROOT1,ROOT2);   } if (!issub) Issub=issubtree (ROOT1-&GT;LEFT,ROOT2);         if (!issub) Issub=issubtree (ROOT1-&GT;RIGHT,ROOT2); } return issub;} </strong>int getheight (treenode* root) {if (root==null) return 0;return getheight (root->left) >getHeight ( root->right)? (GetHeight (root->left) +1):(getheight (root->right) +1);} void Printbylevel (TreeNode *root,int level) {if (root==null| |  LEVEL&LT;0) return;    else if (level==0) {cout<<root->value<< "";     return;  } else {printbylevel (ROOT-&GT;LEFT,LEVEL-1);  Printbylevel (ROOT-&GT;RIGHT,LEVEL-1);    }}int Main () {treenode* root=null,*root2=null;    Insert (&root,20);    Insert (&root,10);    Insert (&root,30);    Insert (&root,15);    Insert (&root,25); insert (&root,35);    Insert (&root,5);    Insert (&root2,10);    Insert (&root2,5);    Insert (&AMP;ROOT2,100);    int high=getheight (root);         for (int i=0;i!=high;i++) {printbylevel (root,i);  cout<<endl;      cout<< "---------" <<endl;    } int High2=getheight (ROOT2);      for (int i=0;i!=high2;i++) {printbylevel (root2,i);  cout<<endl;      cout<< "---------" <<endl;  } cout<<issubtree (Root,root2); }

Time complexity analysis, assuming that the number of nodes in ROOT1 is M.

<span style= "font-family:arial, Helvetica, Sans-serif;" > </span><span style= "font-family:arial, Helvetica, Sans-serif;" >root2 the number of nodes is N. The worst time complexity is O (m*n) </span>

<span style= "font-family:arial, Helvetica, Sans-serif;" The steps of the > algorithm are such:</span>

<span style= "font-family:arial, Helvetica, Sans-serif;" > Traverse tree1 </span>

Find the node that is equal to the Tree2 root node p2

Every time I find P2. Checks if the subtree with P2 as the root node is equal to Tree2, and if it is equal, stops traversing

Assuming that the tree1 has a K node value equal to the root node of the tree1, the actual worst time complexity is O (m+k*n)

So time efficiency is still acceptable.

<span style= "font-family:arial, Helvetica, Sans-serif;" > </span>

How to tell if a binary tree tree is a subtree of another binary tree