Given two binary trees, the task is to check whether the two binary trees is a mirror of each other or not.
Mirror of a Binary Tree: Mirror of a Binary Tree T is another Binary Tree M(T) with left and right children of all non-leaf nodes interchanged.
Trees in the above figure are mirrors of each other.
A recursive solution and an iterative method using inorder traversal to check whether the two binary trees is a mirror of each other or not have been already discussed. In this post a solution using level order traversal has been discussed.
The idea is to use a queue in which two nodes of both the trees which needs to be checked for equality are present together. At each step of level order traversal, get two nodes from the queue, check for their equality and then insert next two children nodes of these nodes which need to be checked for equality. During insertion step, first left child of first tree node and right child of second tree node are inserted. After this right child of first tree node and left child of second tree node are inserted. If at any stage one node is NULL and other is not, then both trees are not a mirror of each other.
Below is the implementation of above approach:
Time complexity: O(N)
- Check if the level order traversal of a Binary Tree results in a palindrome
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- Print a Binary Tree in Vertical Order | Set 3 (Using Level Order Traversal)
- Flatten Binary Tree in order of Level Order Traversal
- Insertion in n-ary tree in given order and Level order traversal
- Reverse Level Order Traversal
- Construct BST from its given level order traversal | Set-2
- Level Order Tree Traversal
- Construct BST from its given level order traversal
- Level order traversal in spiral form
- Zig Zag Level order traversal of a tree using single queue
- Density of Binary Tree using Level Order Traversal
- Reverse Level Order traversal in spiral form
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