Print a Binary Tree in Vertical Order | Set 3 (Using Level Order Traversal)
Given a binary tree, print it vertically. The following example illustrates vertical order traversal.
1 / \ 2 3 / \ / \ 4 5 6 7 \ \ 8 9 The output of print this tree vertically will be: 4 2 1 5 6 3 8 7 9
We have discussed an efficient approach in below post.
Print a Binary Tree in Vertical Order | Set 2 (Hashmap based Method)
The above solution uses preorder traversal and Hashmap to store nodes according to horizontal distances. Since the above approach uses preorder traversal, nodes in a vertical line may not be printed in the same order as they appear in the tree. For example, the above solution prints 12 before 9 in the below tree. See this for a sample run.
1 / \ 2 3 / \ / \ 4 5 6 7 \ / \ 8 10 9 \ 11 \ 12
If we use level order traversal, we can make sure that if a node like 12 comes below in the same vertical line, it is printed after a node like 9 which comes above in the vertical line.
1. To maintain a hash for the branch of each node. 2. Traverse the tree in level order fashion. 3. In level order traversal, maintain a queue which holds, node and its vertical branch. * pop from queue. * add this node's data in vector corresponding to its branch in the hash. * if this node hash left child, insert in the queue, left with branch - 1. * if this node hash right child, insert in the queue, right with branch + 1.
Vertical order traversal is 4 2 1 5 6 3 8 10 7 11 9 12
Time Complexity of the above implementation is O(n Log n). Note that the above implementation uses a map which is implemented using self-balancing BST.
We can reduce the time complexity to O(n) using unordered_map. To print nodes in the desired order, we can have 2 variables denoting min and max horizontal distance. We can simply iterate from min to max horizontal distance and get corresponding values from Map. So it is O(n)
Auxiliary Space: O(n)
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