Given a Binary Tree, the task is to flatten it in order of Level order traversal of the tree. In the flattened binary tree, the left node of all the nodes must be NULL.
Input: 1 / \ 5 2 / \ / \ 6 4 9 3 Output: 1 5 2 6 4 9 3 Input: 1 \ 2 \ 3 \ 4 \ 5 Output: 1 2 3 4 5
Approach: We will solve this problem by simulating the Level order traversal of Binary Tree as follows:
- Create a queue to store the nodes of Binary tree.
- Create a variable “prev” and initialise it by parent node.
- Push left and right children of parent in the queue.
- Apply level order traversal. Lets say “curr” is front most element in queue. Then,
- If ‘curr’ is NULL, continue.
- Else push curr->left and curr->right in the queue
- Set prev = curr
Below is the implementation of the above approach:
Time Complexity: O(N)
Space Complexity: O(N) where N is the size of Binary Tree.
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- Level Order Tree Traversal
- Zig Zag Level order traversal of a tree using single queue
- General Tree (Each node can have arbitrary number of children) Level Order Traversal
- Level Order Successor of a node in Binary Tree
- Level Order Predecessor of a node in Binary Tree
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Improved By : Rajput-Ji