C++ Program to Check Whether an Input Binary Tree is the Sub Tree of the Binary Tree

C++Server Side ProgrammingProgramming

A binary tree is a tree data structure in which each node has at most two children, which are defined as left child and right child.


   function identical(): 
      Take two nodes r1 and r2 as parameter.
      If r1 and r2 is NULL then
         Return true.
      If r1 or r2 is NULL then
         Return false.
      Return (r1->d is equal to r2->d and
         Call function Identical(r1->l, r2->l) and
         Call functions Identical(r1->r, r2->r) );
      function Subtree(node *T, node *S) : 
         if (S == NULL) then
            return true;
         if (T == NULL) then
            return false;
         if (call function Identical(T, S))
            return true;
         return Subtree(T->l, S) or Subtree(T->r, S); 

Example Code

 Live Demo

#include <iostream>
using namespace std;
struct n {
   int d;
   struct n* l;
   struct n* r;
bool Identical(struct n * r1, struct n *r2) {
   if (r1 == NULL && r2 == NULL)
      return true;
   if (r1 == NULL || r2 == NULL)
      return false;
      return (r1->d == r2->d && Identical(r1->l, r2->l) && Identical(r1->r, r2->r) );
bool Subtree(struct n *T, struct n *S) {
   if (S == NULL)
      return true;
   if (T == NULL)
      return false;
   if (Identical(T, S))
      return true;
      return Subtree(T->l, S) || Subtree(T->r, S);
struct n* newN(int d) {
   struct n* nod =
   (struct n*)malloc(sizeof(struct n));
   nod->d = d;
   nod->l = NULL;
   nod->r = NULL;
int main() {
   struct n *T = newN(24);
   T->r = newN(2);
   T->r->r = newN(5);
   T->l = newN(7);
   T->l->l = newN(3);
   T->l->l->r = newN(40);
   T->l->r = newN(6);
   struct n *S = newN(20);
   S->r = newN(5);
   S->l = newN(3);
   S->l->r = newN(50);
   if (Subtree(T, S))
      cout<<"given tree is subtree of Binary tree"<<"\n";
      cout<<"given tree is not subtree of Binary tree"<<"\n";
      struct n *T1 = newN(30);
      T1->r = newN(20);
      T1->r->r = newN(19);
      T1->l = newN(17);
      T1->l->l = newN(4);
      T1->l->l->r = newN(40);
      T1->l->r = newN(15);
      struct n *S1 = newN(17);
      S1->r = newN(15);
      S1->l = newN(4);
      S1->l->r = newN(40);
      if (Subtree(T1, S1))
         cout<<"given tree is subtree of Binary tree";
         cout<<"given tree is not subtree of Binary tree";
         return 0;


given tree is not subtree of Binary tree
given tree is subtree of Binary tree
Published on 02-Apr-2019 13:57:39