Print Top View of Binary Tree

Given an binary tree, print top view of it. Assume, the left and right child of a node makes 45 degree angle with the parent.


For example, Top view of below tree is 2, 1, 3, 6

Top view binary tree


This problem can be easily solved with the help of Hashing. The idea is to create an empty map where each key represents the relative horizontal distance of the node from the root node and value in the map maintains a pair containing node’s value and its level number. Then we do a pre-order traversal of the tree and if current level of a node is less than maximum level seen so far for the same horizontal distance as current node’s or current horizontal distance is seen for the first time, we update the value and the level for current horizontal distance in the map. For each node, we recurse for its left subtree by decreasing horizontal distance and increasing level by 1 and recurse for right subtree by increasing both level and horizontal distance by 1.

Below figure shows horizontal distance and level of each node in above binary tree. The final values in the map will be

horizontal distance -> (node’s value, node’s level)
-1 -> (2, 2)
0 -> (1, 1)
1 -> (3, 2)
2 -> (6, 3)

Horizontal distance vs Level - Binary Tree

C++ implementation –

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The time complexity of above solution is O(nlog(n)) and auxiliary space used by the program is O(n).

Exercise: Modify the solution to print top view of binary tree

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Avishek Dutta

A O(n) solution. The idea is same but the implementation is different.


Hey, the example output looks incorrect.


The condition (|| level < map[dist].second) seems redundant, since the traverse is from lower to deeper level.