Convert binary tree to mirror tree
Given a binary tree, convert this binary tree into its mirror tree.

A binary tree is a tree in which each parent node has at most two children.

Mirror of a Tree: Mirror of a Binary Tree T is another Binary Tree M(T) with left and right children of all non-leaf nodes interchanged.

Note:

1. Make in-place changes, that is, modify the nodes given a binary tree to get the required mirror tree.

Input Format:

The first line of input contains an integer ‘T’ denoting the number of test cases. The first line of input contains the elements of the tree in the level order form separated by a single space. If any node does not have a left or right child, take -1 in its place. Refer to the example below.

Example:

Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place. For example, the input for the tree depicted in the below image would be :

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Explanation :

Level 1 : The root node of the tree is 1 Level 2 : Left child of 1 = 2 Right child of 1 = 3 Level 3 : Left child of 2 = 4 Right child of 2 = null (-1) Left child of 3 = 5 Right child of 3 = 6 Level 4 : Left child of 4 = null (-1) Right child of 4 = 7 Left child of 5 = null (-1) Right child of 5 = null (-1) Left child of 6 = null (-1) Right child of 6 = null (-1) Level 5 : Left child of 7 = null (-1) Right child of 7 = null (-1) The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on. The input ends when all nodes at the last level are null (-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as: 1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format:

For every test case, the inorder traversal of the mirror tree will be printed on a separate line with the elements of the mirror tree in the level order form separated by a single space and -1 in place of NULL left or right child.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraint :

1

Question

Convert binary tree to mirror tree

Given a binary tree, convert this binary tree into its mirror tree.

A binary tree is a tree in which each parent node has at most two children.

Mirror of a Tree: Mirror of a Binary Tree T is another Binary Tree M(T) with left and right children of all non-leaf nodes interchanged.

alt text

Note:

1. Make in-place changes, that is, modify the nodes given a binary tree to get the required mirror tree.

Input Format:

The first line of input contains an integer ‘T’ denoting the number of test cases.

The first line of input contains the elements of the tree in the level order form separated by a single space. If any node does not have a left or right child, take -1 in its place. Refer to the example below.

Example:

Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :

alt text

 1
 2 3
 4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :

Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. 

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format:

For every test case, the inorder traversal of the mirror tree will be printed on a separate line with the elements of the mirror tree in the level order form separated by a single space and -1 in place of NULL left or right child.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraint :

1 <= T <= 100
1 <= N <= 3000
-10^9 <= DATA <= 10^9

Where ‘N’ is the number of nodes in the tree, and 'DATA' denotes data contained in the node of a binary tree.

Time Limit: 1 sec

Accepted Answer

I first clarified the doubts since the question was too vague in start.
Then I told that this problem can be solved with iterative or recursive approach, Interviewer told me to go ahead and code the recursive one.
We discussed time complexity.
Then I implemented iterative one too.

Accepted Answer

Recursive approach.

The idea is to traverse the binary tree and swap the left and right subtrees.

The steps are as follows:

Call mirror function as mirror(node->left) to access the left subtree.
Call mirror function as mirror(node->right) to access the right subtree.
Swap left and right subtrees using:

‘TEMP’ = leftSubtree
leftSubtree = rightSubtree
rightSubtree = ‘TEMP’

Space Complexity: O(n)Explanation:

O(N), Where ‘N’ is the number of nodes in the binary tree.

Since stack takes space for recursive calls of the order ‘N’. Thus the space complexity will be O(N).

Time Complexity: O(n)Explanation:

O(N), Where ‘N’ is the number of nodes in the binary tree.

Since we are recursively traversing through all the nodes of the binary tree. Thus the time complexity will be O(N).

Accepted Answer

Iterative approach

The idea is to do queue-based level order traversal. While doing traversal, swap left and right children of every node.

The steps are as follows:

Create a queue and push the root in the queue.
Run a loop while the queue is not empty and perform the below operations.
a. Store the front element of the queue in a variable and pop it from the queue.
b. Swap the left and right child of the popped element and if they are not NULL, push them in the queue.

Space Complexity: O(n)Explanation:

O(N), Where ‘N’ is the number of nodes in the binary tree.

Since a queue of size ‘N’ is used to store the nodes. Thus the space complexity will be O(N).

Time Complexity: O(n)Explanation:

O(N), Where ‘N’ is the number of nodes in the binary tree.

Since we iterate through all the nodes in the binary tree. Thus the time complexity will be O(N).

Given a binary tree, convert this binary tree into its mirror tree.

A binary tree is a tree in which each parent node has at most two children.

Mirror of a Tree: Mirror of a Binary Tree T is another Binary Tree M(T) with left and right children of all non-leaf nodes interchanged.

Note:

Input Format:

Example:

Explanation :

Note :

Output Format:

Note:

Constraint :

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