Invert Binary Tree - Change to Mirror Tree
Last Updated :
06 Feb, 2025
Given a binary tree, the task is to convert the binary tree to its Mirror tree. Mirror of a Binary Tree T is another Binary Tree M(T) with left and right children of all non-leaf nodes interchanged.
Example 1:
Explanation: In the inverted tree, every non-leaf node has its left and right child interchanged.
Example 2:
Explanation: In the inverted tree, every non-leaf node has its left and right child interchanged.
Recursive Approach - O(n) Time and O(h) Space
The idea is to use recursion to traverse the tree in Post Order (left, right, root) and while traversing each node, swap the left and right subtrees.
C++
//Driver Code Starts
// C++ Program Invert a Binary Tree using Recursive Postorder
#include <iostream>
#include <queue>
using namespace std;
class Node {
public:
int data;
Node *left, *right;
Node(int x) {
data = x;
left = nullptr;
right = nullptr;
}
};
//Driver Code Ends
// function to return the root of inverted tree
void mirror(Node* root) {
if (root == nullptr)
return ;
// Invert the left and right subtree
mirror(root->left);
mirror(root->right);
// Swap the left and right subtree
swap(root->left, root->right);
}
// Print tree as level order
void levelOrder(Node *root) {
if (root == nullptr) {
cout << "N ";
return ;
}
queue<Node*> qq;
qq.push(root);
while (!qq.empty()) {
Node *curr = qq.front();
qq.pop();
if (curr == nullptr) {
cout << "N ";
continue;
}
cout << (curr->data) << " ";
qq.push(curr->left);
qq.push(curr->right);
}
}
//Driver Code Starts
int main() {
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
return 0;
}
//Driver Code Ends
//Driver Code Starts
// C Program Invert a Binary Tree using Recursive Postorder
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
typedef struct Node {
int data;
struct Node* left;
struct Node* right;
} Node;
// Function to create a new node
Node* newNode(int x) {
Node* node = (Node*)malloc(sizeof(Node));
node->data = x;
node->left = NULL;
node->right = NULL;
return node;
}
//Driver Code Ends
// Function to return the root of inverted tree
void mirror(Node* root) {
if (root == NULL)
return ;
// Invert the left and right subtree
mirror(root->left);
mirror(root->right);
// Swap the left and right subtree
struct Node* temp = root->left;
root->left = root->right;
root->right = temp;
}
// Print tree as level order
void levelOrder(Node* root) {
if (root == NULL) {
printf("N ");
return;
}
Node* queue[100];
int front = 0, rear = 0;
queue[rear++] = root;
while (front < rear) {
Node* curr = queue[front++];
if (curr == NULL) {
printf("N ");
continue;
}
printf("%d ", curr->data);
queue[rear++] = curr->left;
queue[rear++] = curr->right;
}
}
//Driver Code Starts
int main() {
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
return 0;
}
//Driver Code Ends
//Driver Code Starts
// Java Program Invert a Binary Tree using Recursive Postorder
import java.util.LinkedList;
import java.util.Queue;
class Node {
int data;
Node left, right;
Node(int x) {
data = x;
left = null;
right = null;
}
}
class GfG {
//Driver Code Ends
// Function to return the root of inverted tree
static void mirror(Node root) {
if (root == null)
return ;
// Invert the left and right subtree
mirror(root.left);
mirror(root.right);
// Swap the left and right subtree
Node temp = root.left;
root.left = root.right;
root.right = temp;
}
// Print tree as level order
static void levelOrder(Node root) {
if (root == null) {
System.out.print("N ");
return;
}
Queue<Node> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
Node curr = queue.poll();
if (curr == null) {
System.out.print("N ");
continue;
}
System.out.print(curr.data + " ");
queue.add(curr.left);
queue.add(curr.right);
}
}
//Driver Code Starts
public static void main(String[] args) {
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
}
}
//Driver Code Ends
#Driver Code Starts
# Python Program Invert a Binary Tree using Recursive Postorder
from collections import deque
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
#Driver Code Ends
# Function to return the root of inverted tree
def mirror(root):
if root is None:
return
# Invert the left and right subtree
mirror(root.left)
mirror(root.right)
# Swap the left and right subtree
temp = root.left
root.left = root.right
root.right = temp
# Print tree as level order
def levelOrder(root):
if root is None:
print("N ", end="")
return
queue = deque([root])
while queue:
curr = queue.popleft()
if curr is None:
print("N ", end="")
continue
print(curr.data, end=" ")
queue.append(curr.left)
queue.append(curr.right)
#Driver Code Starts
if __name__ == "__main__":
# Input Tree:
# 1
# / \
# 2 3
# / \
# 4 5
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
mirror(root)
# Mirror Tree:
# 1
# / \
# 3 2
# / \
# 5 4
levelOrder(root)
#Driver Code Ends
//Driver Code Starts
// C# Program Invert a Binary Tree using Recursive Postorder
using System;
using System.Collections.Generic;
class Node {
public int data;
public Node left, right;
public Node(int x) {
data = x;
left = null;
right = null;
}
}
class GfG {
//Driver Code Ends
// Function to return the root of inverted tree
static void mirror(Node root) {
if (root == null)
return ;
// Invert the left and right subtree
mirror(root.left);
mirror(root.right);
// Swap the left and right subtree
Node temp = root.left;
root.left = root.right;
root.right = temp;
}
// Print tree as level order
static void levelOrder(Node root) {
if (root == null) {
Console.Write("N ");
return;
}
Queue<Node> queue = new Queue<Node>();
queue.Enqueue(root);
while (queue.Count > 0) {
Node curr = queue.Dequeue();
if (curr == null) {
Console.Write("N ");
continue;
}
Console.Write(curr.data + " ");
queue.Enqueue(curr.left);
queue.Enqueue(curr.right);
}
}
//Driver Code Starts
static void Main(string[] args) {
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
}
}
//Driver Code Ends
//Driver Code Starts
// JavaScript Program Invert a Binary Tree using Recursive Postorder
class Node {
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
//Driver Code Ends
// Function to return the root of inverted tree
function mirror(root) {
if (root === null)
return;
// Invert the left and right subtree
mirror(root.left);
mirror(root.right);
// Swap the left and right subtree
let temp = root.left;
root.left = root.right;
root.right = temp;
}
// Print tree as level order
function levelOrder(root) {
if (root === null) {
process.stdout.write("N ");
return;
}
let queue = [];
queue.push(root);
while (queue.length > 0) {
let curr = queue.shift();
if (curr === null) {
process.stdout.write("N ");
continue;
}
process.stdout.write(curr.data + " ");
queue.push(curr.left);
queue.push(curr.right);
}
}
//Driver Code Starts
// Driver Code
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
//Driver Code Ends
Output1 3 2 N N 5 4 N N N N
Time Complexity: O(n), Visiting all the nodes of the tree of size n.
Auxiliary Space: O(h), where h is the height of binary tree.
Iterative Approach - O(n) Time and O(n) Space
The idea is to perform Level Order Traversal using a queue to store the nodes whose left and right pointers need to be swapped. Start with the root node. Till queue is not empty, remove the node from the front, swap its left and right child and push both of the children into the queue. After iterating over all the nodes, we will get the mirror tree.
C++
//Driver Code Starts
// C++ Program Invert a Binary Tree using Iterative Level Order
#include <iostream>
#include <queue>
using namespace std;
struct Node {
int data;
Node *left, *right;
Node(int x) {
data = x;
left = nullptr;
right = nullptr;
}
};
//Driver Code Ends
// function to return the root of inverted tree
void mirror(Node* root) {
if (root == nullptr)
return ;
queue<Node*> q;
q.push(root);
// Traverse the tree, level by level
while(!q.empty()) {
Node* curr = q.front();
q.pop();
// Swap the left and right subtree
swap(curr->left, curr->right);
// Push the left and right node to the queue
if(curr->left != nullptr)
q.push(curr->left);
if(curr->right != nullptr)
q.push(curr->right);
}
}
// Print tree as level order
void levelOrder(Node *root) {
if (root == nullptr) {
cout << "N ";
return ;
}
queue<Node*> qq;
qq.push(root);
while (!qq.empty()) {
Node *curr = qq.front();
qq.pop();
if (curr == nullptr) {
cout << "N ";
continue;
}
cout << (curr->data) << " ";
qq.push(curr->left);
qq.push(curr->right);
}
}
//Driver Code Starts
int main() {
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
return 0;
}
//Driver Code Ends
//Driver Code Starts
// C Program Invert a Binary Tree using Iterative Level Order
#include <stdio.h>
#include <stdlib.h>
// Node structure
typedef struct Node {
int data;
struct Node* left;
struct Node* right;
} Node;
// Function to create a new node
Node* newNode(int x) {
Node* node = (Node*)malloc(sizeof(Node));
node->data = x;
node->left = NULL;
node->right = NULL;
return node;
}
//Driver Code Ends
// Function to return the root of inverted tree
void mirror(Node* root) {
if (root == NULL)
return;
Node* queue[100];
int front = 0, rear = 0;
queue[rear++] = root;
// Traverse the tree, level by level
while (front < rear) {
Node* curr = queue[front++];
// Swap the left and right subtree
Node* temp = curr->left;
curr->left = curr->right;
curr->right = temp;
// Push the left and right node to the queue
if (curr->left != NULL)
queue[rear++] = curr->left;
if (curr->right != NULL)
queue[rear++] = curr->right;
}
}
// Print tree as level order
void levelOrder(Node* root) {
if (root == NULL) {
printf("N ");
return;
}
Node* queue[100];
int front = 0, rear = 0;
queue[rear++] = root;
while (front < rear) {
Node* curr = queue[front++];
if (curr == NULL) {
printf("N ");
continue;
}
printf("%d ", curr->data);
queue[rear++] = curr->left;
queue[rear++] = curr->right;
}
}
//Driver Code Starts
int main() {
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
return 0;
}
//Driver Code Ends
//Driver Code Starts
// Java Program Invert a Binary Tree using Iterative Level Order
import java.util.LinkedList;
import java.util.Queue;
class Node {
int data;
Node left, right;
Node(int x) {
data = x;
left = null;
right = null;
}
}
class GfG {
//Driver Code Ends
// Function to return the root of inverted tree
static void mirror(Node root) {
if (root == null)
return;
Queue<Node> queue = new LinkedList<>();
queue.add(root);
// Traverse the tree, level by level
while (!queue.isEmpty()) {
Node curr = queue.poll();
// Swap the left and right subtree
Node temp = curr.left;
curr.left = curr.right;
curr.right = temp;
// Push the left and right node to the queue
if (curr.left != null)
queue.add(curr.left);
if (curr.right != null)
queue.add(curr.right);
}
}
// Print tree as level order
static void levelOrder(Node root) {
if (root == null) {
System.out.print("N ");
return;
}
Queue<Node> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
Node curr = queue.poll();
if (curr == null) {
System.out.print("N ");
continue;
}
System.out.print(curr.data + " ");
queue.add(curr.left);
queue.add(curr.right);
}
}
//Driver Code Starts
public static void main(String[] args) {
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
}
}
//Driver Code Ends
#Driver Code Starts
# Python Program Invert a Binary Tree using Iterative Level Order
from collections import deque
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
#Driver Code Ends
# Function to return the root of inverted tree
def mirror(root):
if root is None:
return
queue = deque([root])
# Traverse the tree, level by level
while queue:
curr = queue.popleft()
# Swap the left and right subtree
curr.left, curr.right = curr.right, curr.left
# Push the left and right node to the queue
if curr.left:
queue.append(curr.left)
if curr.right:
queue.append(curr.right)
# Print tree as level order
def levelOrder(root):
if root is None:
print("N ", end="")
return
queue = deque([root])
while queue:
curr = queue.popleft()
if curr is None:
print("N ", end="")
continue
print(curr.data, end=" ")
queue.append(curr.left)
queue.append(curr.right)
#Driver Code Starts
if __name__ == "__main__":
# Input Tree:
# 1
# / \
# 2 3
# / \
# 4 5
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
mirror(root)
# Mirror Tree:
# 1
# / \
# 3 2
# / \
# 5 4
levelOrder(root)
#Driver Code Ends
//Driver Code Starts
// C# Program Invert a Binary Tree using Iterative Level Order
using System;
using System.Collections.Generic;
class Node {
public int data;
public Node left, right;
public Node(int x) {
data = x;
left = null;
right = null;
}
}
class GfG {
//Driver Code Ends
// Function to return the root of inverted tree
static void mirror(Node root) {
if (root == null)
return;
Queue<Node> queue = new Queue<Node>();
queue.Enqueue(root);
// Traverse the tree, level by level
while (queue.Count > 0) {
Node curr = queue.Dequeue();
// Swap the left and right subtree
Node temp = curr.left;
curr.left = curr.right;
curr.right = temp;
// Push the left and right node to the queue
if (curr.left != null)
queue.Enqueue(curr.left);
if (curr.right != null)
queue.Enqueue(curr.right);
}
}
// Print tree as level order
static void levelOrder(Node root) {
if (root == null) {
Console.Write("N ");
return;
}
Queue<Node> queue = new Queue<Node>();
queue.Enqueue(root);
while (queue.Count > 0) {
Node curr = queue.Dequeue();
if (curr == null) {
Console.Write("N ");
continue;
}
Console.Write(curr.data + " ");
queue.Enqueue(curr.left);
queue.Enqueue(curr.right);
}
}
//Driver Code Starts
static void Main(string[] args) {
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
}
}
//Driver Code Ends
//Driver Code Starts
// JavaScript Program Invert a Binary Tree using Iterative Level Order
class Node {
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
//Driver Code Ends
// Function to return the root of inverted tree
function mirror(root) {
if (root === null)
return;
let queue = [];
queue.push(root);
// Traverse the tree, level by level
while (queue.length > 0) {
let curr = queue.shift();
// Swap the left and right subtree
[curr.left, curr.right] = [curr.right, curr.left];
// Push the left and right node to the queue
if (curr.left)
queue.push(curr.left);
if (curr.right)
queue.push(curr.right);
}
}
// Print tree as level order
function levelOrder(root) {
if (root === null) {
process.stdout.write("N ");
return;
}
let queue = [];
queue.push(root);
while (queue.length > 0) {
let curr = queue.shift();
if (curr === null) {
process.stdout.write("N ");
continue;
}
process.stdout.write(curr.data + " ");
queue.push(curr.left);
queue.push(curr.right);
}
}
//Driver Code Starts
// Driver Code
// Input Tree:
// 1
// / \
// 2 3
// / \
// 4 5
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
mirror(root);
// Mirror Tree:
// 1
// / \
// 3 2
// / \
// 5 4
levelOrder(root);
//Driver Code Ends
Output1 3 2 N N 5 4 N N N N
Time Complexity: O(n), fro traversing over the tree of size n.
Auxiliary Space: O(n), used by queue to store the nodes of the tree.
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