Check if two trees have same structure

Given two binary trees. The task is to write a program to check if the two trees are identical in structure.

In the above figure both of the trees, Tree1 and Tree2 are identical in structure. That is, they have the same structure.



Note: This problem is different from Check if two trees are identical as here we need to compare only the structures of the two trees and not the values at their nodes.

The idea is to traverse both trees simultaneously following the same paths and keep checking if a node exists for both the trees or not.

Algorithm:

  1. If both trees are empty then return 1.
  2. Else If both trees are non-empty:
    • Check left subtrees recursively i.e., call isSameStructure(tree1->left_subtree, tree2->left_subtree)
    • Check right subtrees recursively i.e., call isSameStructure(tree1->right_subtree, tree2->right_subtree)
    • If the value returned in above two steps are true then return 1.
  3. Else return 0 (one is empty and other is not).

Below is the implementation of above algorithm:

C++

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// C++ program to check if two trees have 
// same structure
#include <iostream>
using namespace std;
  
// A binary tree node has data, pointer to left child
// and a pointer to right child 
struct Node
{
    int data;
    struct Node* left;
    struct Node* right;
};
  
// Helper function that allocates a new node with the
// given data and NULL left and right pointers. 
Node* newNode(int data)
{
    Node* node = new Node;
    node->data = data;
    node->left = NULL;
    node->right = NULL;
  
    return(node);
}
  
// Function to check if two trees have same 
// structure
int isSameStructure(Node* a, Node* b)
{
    // 1. both empty 
    if (a==NULL && b==NULL)
        return 1;
  
    // 2. both non-empty -> compare them 
    if (a!=NULL && b!=NULL)
    {
        return
        (
            isSameStructure(a->left, b->left) &&
            isSameStructure(a->right, b->right)
        );
    }
      
    // 3. one empty, one not -> false 
    return 0;
  
// Driver code
int main()
{
    Node *root1 = newNode(10);
    Node *root2 = newNode(100);
    root1->left = newNode(7);
    root1->right = newNode(15);
    root1->left->left = newNode(4);
    root1->left->right = newNode(9); 
    root1->right->right = newNode(20);
  
    root2->left = newNode(70);
    root2->right = newNode(150);
    root2->left->left = newNode(40);
    root2->left->right = newNode(90); 
    root2->right->right = newNode(200);
  
    if (isSameStructure(root1, root2))
        printf("Both trees have same structure");
    else
        printf("Trees do not have same structure");
  
    return 0;
}

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Java

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// Java program to check if two trees have 
// same structure 
class GFG
{
      
// A binary tree node has data, 
// pointer to left child and 
// a pointer to right child 
static class Node 
    int data; 
    Node left; 
    Node right; 
}; 
  
// Helper function that allocates a new node 
// with the given data and null left 
// and right pointers. 
static Node newNode(int data) 
    Node node = new Node(); 
    node.data = data; 
    node.left = null
    node.right = null
  
    return(node); 
  
// Function to check if two trees 
// have same structure 
static boolean isSameStructure(Node a, Node b) 
    // 1. both empty 
    if (a == null && b == null
        return true
  
    // 2. both non-empty . compare them 
    if (a != null && b != null
    
        return
        
            isSameStructure(a.left, b.left) && 
            isSameStructure(a.right, b.right) 
        ); 
    
      
    // 3. one empty, one not . false 
    return false
  
// Driver code 
public static void main(String args[])
    Node root1 = newNode(10); 
    Node root2 = newNode(100); 
    root1.left = newNode(7); 
    root1.right = newNode(15); 
    root1.left.left = newNode(4); 
    root1.left.right = newNode(9); 
    root1.right.right = newNode(20); 
  
    root2.left = newNode(70); 
    root2.right = newNode(150); 
    root2.left.left = newNode(40); 
    root2.left.right = newNode(90); 
    root2.right.right = newNode(200); 
  
    if (isSameStructure(root1, root2)) 
        System.out.printf("Both trees have same structure"); 
    else
        System.out.printf("Trees do not have same structure"); 
}
  
// This code is contributed by Arnab Kundu

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C#

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// C# program to check if two trees 
// have same structure 
using System;
  
class GFG
{
      
// A binary tree node has data, 
// pointer to left child and 
// a pointer to right child 
public class Node 
    public int data; 
    public Node left; 
    public Node right; 
}; 
  
// Helper function that allocates a new node 
// with the given data and null left 
// and right pointers. 
static Node newNode(int data) 
    Node node = new Node(); 
    node.data = data; 
    node.left = null
    node.right = null
  
    return(node); 
  
// Function to check if two trees 
// have same structure 
static Boolean isSameStructure(Node a,
                               Node b) 
    // 1. both empty 
    if (a == null && b == null
        return true
  
    // 2. both non-empty . compare them 
    if (a != null && b != null
    
        return
        
            isSameStructure(a.left, b.left) && 
            isSameStructure(a.right, b.right) 
        ); 
    
      
    // 3. one empty, one not . false 
    return false
  
// Driver code 
public static void Main(String []args)
    Node root1 = newNode(10); 
    Node root2 = newNode(100); 
    root1.left = newNode(7); 
    root1.right = newNode(15); 
    root1.left.left = newNode(4); 
    root1.left.right = newNode(9); 
    root1.right.right = newNode(20); 
  
    root2.left = newNode(70); 
    root2.right = newNode(150); 
    root2.left.left = newNode(40); 
    root2.left.right = newNode(90); 
    root2.right.right = newNode(200); 
  
    if (isSameStructure(root1, root2)) 
        Console.Write("Both trees have " +  
                      "same structure"); 
    else
        Console.Write("Trees do not have" +
                      " same structure"); 
}
  
// This code is contributed by Rajput-Ji

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Output:

Both trees have same structure


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Improved By : andrew1234, Rajput-Ji