# Check if two trees are structurally similar or not

In this article, we are going to see **how to check two trees are structurally similar or not?**

Submitted by Radib Kar, on August 07, 2020

In this article, we going to see given two Tree roots how we can find whether** two trees are structurally similar or not.**

## When we can say two trees are structurally similar?

For two trees are said to be **structurally similar if** the roots have the same structural subtrees. No matter what are the values of the roots, that doesn't have any impact on the structural similarity. Previously, we saw how to check if two trees are identical or not? The only difference in structural similarity is that we don't care about the root values. So, if we discard that root value checking, then we can use the same function ** identical() **here too.

Say if ** structurally_similar()** is a function that takes two tree nodes and returns true or false based on where they are

**structurally similar**or not, then based on the above inference we can write like below

** structurally_similar** (root1, root2) returns

**(root1->left,root2->left) &&**

*structurally_similar***root1->right,root2->right))**

*structurally_similar*That means it returns true only if both their left subtrees are structurally similar and both their right subtrees are structurally similar. Otherwise, it returns false.

To check whether both their left and right subtrees are equal or not we used the same function recursively.

**Recursive function to check**

Returns true if two trees are structurally same, otherwise false bool structurally_identical(root1, root2){ /////base cases///// if (both roots are NULL) then they are structurally identical, so return true; if (one root is NULL & other is not) then they are not structurally identical, so return false; //no check for root values //recursively check both their subtrees return structurally_identical (root1->left,root2->left) && structurally_identical(root1->right,root2->right); }

**Example**

**1) When two trees are structurally identical**

Below is the example where both the trees are structurally identical.

If we run a dry-run on the above tree. Then it starts with calling structurally_identical (1,10)

structurally_identical(1,10) Since no root is NULL, it doesn't enter the base cases //no checking for root value So, it now checks for both subtree, call to structurally_identical (1->left, 10->left) && structurally_identical (1->right, 10->right) structurally_identical (1->left, 10->left): structurally_identical (2, 20) Since no root is NULL, it doesn't enter the base case //no checking for root value So, it now checks for both subtree, call to structurally_identical (2->left, 20->left) && structurally_identical (2->right, 20->right) structurally_identical (2->left, 20->left): structurally_identical (4,40) Since no root is NULL, it doesn't enter the base cases //no checking for root value So, it now checks for both subtree, call to structurally_identical (4->left, 40->left) && structurally_identical (4->right, 40->right) Since, structurally_identical (4->left, 40->left) returns true as both roots are NULL. Same way, structurally_identical (4->right, 40->right) returns true So, the entire function structurally_identical (2->left,20->left) returns true, that means the left subtree of root with value 2 & root with value 20 are structurally identical Similarly, We can execute structurally_identical (2->right, 20->right) and that will return true also. So now structurally_identical (1->left, 10->left) returns true that means left subtree of our original roots are structurally identical. We can find right subtrees are also structurally identical, Hence both trees are structurally identical/similar.

**2) When the two trees are not structurally identical**

Below is the example where the trees are not structurally identical.

You can run a dry run of course, but since the length of the article will increase, I am skipping. But to point out why the above trees are not structurally identical and how our algorithm with detect

We can say that at the stage of calling structurally_identical(3,3) it will return false since the left child of 3 is 5 for the first tree whereas it's NULL for the second tree. So it will hit the second base case and return false. Once a function call returns false that will bubble up to the first function call and ultimately lead to an output ** false**.

**C Implementation:**

#include <stdio.h> #include <stdlib.h> struct tree { int val; struct tree* left; struct tree* right; }; typedef struct tree TreeNode; TreeNode* newTree(int data) { // Allocate memory for new node TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode)); // Assign data to this node val root->val = data; // Initialize left and right children as NULL root->left = NULL; root->right = NULL; return (root); } int structurally_identical(TreeNode* root1, TreeNode* root2) { /////base cases///// //1.both roots are NULL then they are structurally identical if (root1 == NULL && root2 == NULL) return 1; //2.one root is NULL & other is not, //then they are not structurally identical if (root1 == NULL || root2 == NULL) return 0; //no check for root values //recursively check both their subtrees return structurally_identical(root1->left, root2->left) && structurally_identical(root1->right, root2->right); } int main() { //building the trees like example 1 TreeNode* root1 = newTree(1); root1->left = newTree(2); root1->right = newTree(3); root1->left->left = newTree(4); root1->right->left = newTree(5); TreeNode* root2 = newTree(10); root2->left = newTree(20); root2->right = newTree(30); root2->left->left = newTree(40); root2->right->left = newTree(50); if (structurally_identical(root1, root2)) printf("Both the trees in example1 are structurally identical\n"); else printf("Both the trees in example1 are not structurally identical\n"); //building the trees like example 2 TreeNode* root3 = newTree(1); root3->left = newTree(2); root3->right = newTree(3); root3->left->left = newTree(4); root3->right->left = newTree(5); TreeNode* root4 = newTree(10); root4->left = newTree(20); root4->right = newTree(30); root4->left->left = newTree(40); root4->right->right = newTree(50); if (structurally_identical(root3, root4)) printf("Both the trees in example2 are structurally identical\n"); else printf("Both the trees in example2 are not structurally identical\n"); return 0; }

**Output:**

Both the trees in example1 are structurally identical Both the trees in example2 are not structurally identical

**C++ implementation:**

#include <bits/stdc++.h> using namespace std; // tree node is defined class TreeNode { public: int val; TreeNode* left; TreeNode* right; TreeNode(int data) { val = data; left = NULL; right = NULL; } }; bool structurally_identical(TreeNode* root1, TreeNode* root2) { /////base cases///// //1.both roots are NULL then they are structurally identical if (root1 == NULL && root2 == NULL) return true; //2.one root is NULL & other is not, //then they are not structurally identical if (root1 == NULL || root2 == NULL) return false; //no check for root values //recursively check both their subtrees return structurally_identical(root1->left, root2->left) && structurally_identical(root1->right, root2->right); } int main() { //building the trees like example 1 TreeNode* root1 = new TreeNode(1); root1->left = new TreeNode(2); root1->right = new TreeNode(3); root1->left->left = new TreeNode(4); root1->right->left = new TreeNode(5); TreeNode* root2 = new TreeNode(10); root2->left = new TreeNode(20); root2->right = new TreeNode(30); root2->left->left = new TreeNode(40); root2->right->left = new TreeNode(50); if (structurally_identical(root1, root2)) cout << "Both the trees in example1 are structurally identical\n"; else cout << "Both the trees in example1 are not structurally identical\n"; //building the trees like example 2 TreeNode* root3 = new TreeNode(1); root3->left = new TreeNode(2); root3->right = new TreeNode(3); root3->left->left = new TreeNode(4); root3->right->left = new TreeNode(5); TreeNode* root4 = new TreeNode(10); root4->left = new TreeNode(20); root4->right = new TreeNode(30); root4->left->left = new TreeNode(40); root4->right->right = new TreeNode(50); if (structurally_identical(root3, root4)) cout << "Both the trees in example2 are structurally identical\n"; else cout << "Both the trees in example2 are not structurally identical\n"; return 0; }

**Output:**

Both the trees in example1 are structurally identical Both the trees in example2 are not structurally identical

TOP Interview Coding Problems/Challenges

- Run-length encoding (find/print frequency of letters in a string)
- Sort an array of 0's, 1's and 2's in linear time complexity
- Checking Anagrams (check whether two string is anagrams or not)
- Relative sorting algorithm
- Finding subarray with given sum
- Find the level in a binary tree with given sum K
- Check whether a Binary Tree is BST (Binary Search Tree) or not
- 1[0]1 Pattern Count
- Capitalize first and last letter of each word in a line
- Print vertical sum of a binary tree
- Print Boundary Sum of a Binary Tree
- Reverse a single linked list
- Greedy Strategy to solve major algorithm problems
- Job sequencing problem
- Root to leaf Path Sum
- Exit Point in a Matrix
- Find length of loop in a linked list
- Toppers of Class
- Print All Nodes that don't have Sibling
- Transform to Sum Tree
- Shortest Source to Destination Path

Comments and Discussions

**Ad:**
Are you a blogger? Join our Blogging forum.