# Check for Symmetric Binary Tree (Iterative Approach)

Given a binary tree, check whether it is a mirror of itself without recursion.

Examples:

Input :

1
/   \
2     2
/ \   / \
3   4 4   3

Output : Symmetric

Input :

1
/ \
2   2
\   \
3    3

Output : Not Symmetric
Recommended Practice

We have discussed recursive approach to solve this problem in below post :
Symmetric Tree (Mirror Image of itself)
In this post, iterative approach is discussed. We use Queue here. Note that for a symmetric tree elements at every level are palindromic. In example 2, at the leaf level, the elements are not palindromic.
In other words,

1. The left child of left subtree = right child of right subtree.
2. The right child of left subtree = left child of right subtree.

If we insert the left child of left subtree first followed by right child of the right subtree in the queue, we only need to ensure that these are equal.
Similarly, If we insert the right child of left subtree followed by left child of the right subtree in the queue, we again need to ensure that these are equal.

Below is the implementation based on above idea.

## C++

 // C++ program to check if a given Binary // Tree is symmetric or not #include using namespace std;   // A Binary Tree Node struct Node {     int key;     struct Node* left, *right; };   // Utility function to create new Node Node *newNode(int key) {     Node *temp = new Node;     temp->key = key;     temp->left = temp->right = NULL;     return (temp); }   // Returns true if a tree is symmetric // i.e. mirror image of itself bool isSymmetric(struct Node* root) {     if(root == NULL)         return true;           // If it is a single tree node, then     // it is a symmetric tree.     if(!root->left && !root->right)         return true;           queue q;           // Add root to queue two times so that     // it can be checked if either one child     // alone is NULL or not.     q.push(root);     q.push(root);           // To store two nodes for checking their     // symmetry.     Node* leftNode, *rightNode;           while(!q.empty()){                   // Remove first two nodes to check         // their symmetry.         leftNode = q.front();         q.pop();                   rightNode = q.front();         q.pop();                   // if both left and right nodes         // exist, but have different         // values--> inequality, return false         if(leftNode->key != rightNode->key){         return false;         }                   // Push left child of left subtree node         // and right child of right subtree         // node in queue.         if(leftNode->left && rightNode->right){             q.push(leftNode->left);             q.push(rightNode->right);         }                   // If only one child is present alone         // and other is NULL, then tree         // is not symmetric.         else if (leftNode->left || rightNode->right)         return false;                   // Push right child of left subtree node         // and left child of right subtree node         // in queue.         if(leftNode->right && rightNode->left){             q.push(leftNode->right);             q.push(rightNode->left);         }                   // If only one child is present alone         // and other is NULL, then tree         // is not symmetric.         else if(leftNode->right || rightNode->left)         return false;     }           return true; }   // Driver program int main() {     // Let us construct the Tree shown in     // the above figure     Node *root = newNode(1);     root->left = newNode(2);     root->right = newNode(2);     root->left->left = newNode(3);     root->left->right = newNode(4);     root->right->left = newNode(4);     root->right->right = newNode(3);       if(isSymmetric(root))         cout << "The given tree is Symmetric";     else         cout << "The given tree is not Symmetric";     return 0; }   // This code is contributed by Nikhil jindal.

## Java

 // Iterative Java program to check if // given binary tree symmetric import java.util.* ;   public class BinaryTree {     Node root;     static class Node     {         int val;         Node left, right;         Node(int v)         {             val = v;             left = null;             right = null;         }     }       /* constructor to initialise the root */     BinaryTree(Node r) { root = r; }       /* empty constructor */     BinaryTree()  {  }         /* function to check if the tree is Symmetric */     public boolean isSymmetric(Node root)     {         /* This allows adding null elements to the queue */         Queue q = new LinkedList();           /* Initially, add left and right nodes of root */         q.add(root.left);         q.add(root.right);           while (!q.isEmpty())         {             /* remove the front 2 nodes to               check for equality */             Node tempLeft = q.remove();             Node tempRight = q.remove();               /* if both are null, continue and check                for further elements */             if (tempLeft==null && tempRight==null)                 continue;               /* if only one is null---inequality, return false */             if ((tempLeft==null && tempRight!=null) ||                 (tempLeft!=null && tempRight==null))                 return false;               /* if both left and right nodes exist, but                have different values-- inequality,                return false*/             if (tempLeft.val != tempRight.val)                 return false;               /* Note the order of insertion of elements                to the queue :                1) left child of left subtree                2) right child of right subtree                3) right child of left subtree                4) left child of right subtree */             q.add(tempLeft.left);             q.add(tempRight.right);             q.add(tempLeft.right);             q.add(tempRight.left);         }           /* if the flow reaches here, return true*/         return true;     }       /* driver function to test other functions */     public static void main(String[] args)     {         Node n = new Node(1);         BinaryTree bt = new BinaryTree(n);         bt.root.left = new Node(2);         bt.root.right = new Node(2);         bt.root.left.left = new Node(3);         bt.root.left.right = new Node(4);         bt.root.right.left = new Node(4);         bt.root.right.right = new Node(3);           if (bt.isSymmetric(bt.root))             System.out.println("The given tree is Symmetric");         else             System.out.println("The given tree is not Symmetric");     } }

## Python3

 # Python3 program to program to check if a # given Binary Tree is symmetric or not   # Helper function that allocates a new # node with the given data and None # left and right pairs.                                     class newNode:       # Constructor to create a new node     def __init__(self, key):         self.key = key         self.left = None         self.right = None   # function to check if a given # Binary Tree is symmetric or not def isSymmetric( root) :       # if tree is empty     if (root == None) :         return True           # If it is a single tree node,     # then it is a symmetric tree.     if(not root.left and not root.right):         return True           q = []               # Add root to queue two times so that     # it can be checked if either one     # child alone is NULL or not.     q.append(root)     q.append(root)           # To store two nodes for checking     # their symmetry.     leftNode = 0     rightNode = 0           while(not len(q)):                   # Remove first two nodes to         # check their symmetry.         leftNode = q[0]         q.pop(0)                   rightNode = q[0]         q.pop(0)                   # if both left and right nodes         # exist, but have different         # values-. inequality, return False         if(leftNode.key != rightNode.key):             return False                   # append left child of left subtree         # node and right child of right          # subtree node in queue.         if(leftNode.left and rightNode.right) :             q.append(leftNode.left)             q.append(rightNode.right)                   # If only one child is present         # alone and other is NULL, then         # tree is not symmetric.         elif (leftNode.left or rightNode.right) :             return False                   # append right child of left subtree         # node and left child of right subtree         # node in queue.         if(leftNode.right and rightNode.left):             q.append(leftNode.right)             q.append(rightNode.left)                   # If only one child is present         # alone and other is NULL, then         # tree is not symmetric.         elif(leftNode.right or rightNode.left):             return False           return True           # Driver Code if __name__ == '__main__':           # Let us construct the Tree     # shown in the above figure     root = newNode(1)     root.left = newNode(2)     root.right = newNode(2)     root.left.left = newNode(3)     root.left.right = newNode(4)     root.right.left = newNode(4)     root.right.right = newNode(3)     if (isSymmetric(root)) :         print("The given tree is Symmetric")     else:         print("The given tree is not Symmetric")   # This code is contributed by # Shubham Singh(SHUBHAMSINGH10)

## C#

 // Iterative C# program to check if // given binary tree symmetric using System; using System.Collections.Generic;   public class BinaryTree {     public Node root;     public class Node     {         public int val;         public Node left, right;         public Node(int v)         {             val = v;             left = null;             right = null;         }     }       /* constructor to initialise the root */     BinaryTree(Node r) { root = r; }       /* empty constructor */     BinaryTree() { }         /* function to check if the tree is Symmetric */     public bool isSymmetric(Node root)     {         /* This allows adding null elements to the queue */         Queue q = new Queue();           /* Initially, add left and right nodes of root */         q.Enqueue(root.left);         q.Enqueue(root.right);           while (q.Count!=0)         {             /* remove the front 2 nodes to             check for equality */             Node tempLeft = q.Dequeue();             Node tempRight = q.Dequeue();               /* if both are null, continue and check             for further elements */             if (tempLeft==null && tempRight==null)                 continue;               /* if only one is null---inequality, return false */             if ((tempLeft==null && tempRight!=null) ||                 (tempLeft!=null && tempRight==null))                 return false;               /* if both left and right nodes exist, but             have different values-- inequality,             return false*/             if (tempLeft.val != tempRight.val)                 return false;               /* Note the order of insertion of elements             to the queue :             1) left child of left subtree             2) right child of right subtree             3) right child of left subtree             4) left child of right subtree */             q.Enqueue(tempLeft.left);             q.Enqueue(tempRight.right);             q.Enqueue(tempLeft.right);             q.Enqueue(tempRight.left);         }           /* if the flow reaches here, return true*/         return true;     }       /* driver code */     public static void Main(String[] args)     {         Node n = new Node(1);         BinaryTree bt = new BinaryTree(n);         bt.root.left = new Node(2);         bt.root.right = new Node(2);         bt.root.left.left = new Node(3);         bt.root.left.right = new Node(4);         bt.root.right.left = new Node(4);         bt.root.right.right = new Node(3);           if (bt.isSymmetric(bt.root))             Console.WriteLine("The given tree is Symmetric");         else             Console.WriteLine("The given tree is not Symmetric");     } }   // This code is contributed by PrinciRaj1992

## Javascript



Output

The given tree is Symmetric

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(h), where h is the height of the tree.

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