Top View of Binary Tree
A medium Binary Trees interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Binary Trees
- Sheets
- 2
- Core for
- 4 roles
The problem
Given the root of a binary tree, return the values of the nodes visible from the top when looking at the tree from above. A node is visible from the top if no node above it in the same vertical column has a smaller row.
Example 1
- Input
- [1,2,3,4,5,6,7]
- Output
- [4,2,1,3,7]
- Why
- Looking from above, the topmost node in each vertical column is visible: 4 at column -2, 2 at -1, 1 at 0, 3 at 1, 7 at 2.
Example 2
- Input
- [1,2,3,null,5,null,7]
- Output
- [2,1,3,7]
- Why
- Nodes visible from top: 2 at column -1, 1 at column 0, 3 at column 1, 7 at column 2.
Example 3
- Input
- [1]
- Output
- [1]
- Why
- The single root node is visible.
Constraints
- The number of nodes is in the range [1, 1000].
- -1000 <= Node.val <= 1000
How to think about it
Updated 2026-09-29The top view contains exactly one node per vertical column: the first node encountered at that horizontal distance. Breadth-first search visits nodes level by level from top to bottom, which means the very first time a column coordinate is visited in BFS, that node is guaranteed to be the top-view node.
Approaches, worst first
DFS with depth map
time O(n log n) · space O(n)
Pass `(col, depth)` recursively. In a map of `col -> (val, depth)`, update an entry only if the current depth is strictly smaller than the existing recorded depth. Extract columns in sorted order after tree traversal completes.
BFS with first-arrival recordingWrite this one
time O(n) · space O(n)
Queue tuples of `(node, col)`. Keep a hash map of column to value. Because BFS descends row by row, if `map` does not contain `col`, record `node.val`. Track min and max column to emit the final array directly without sorting keys.
Where people lose marks · 3
- Using DFS and blindly recording the first visited node for a column without comparing depths; DFS dives deep along the left before visiting shallower nodes in other branches.
- Overwriting earlier recorded entries in BFS; subsequent arrivals on the same column are deeper and therefore obscured from above.
- Sorting map keys at the end when tracking `minCol` and `maxCol` allows O(n) array assembly.
The theory behind it
Binary Trees — the ground this problem stands on. All Binary Trees problems
What Binary Trees is
A binary tree is a branching data structure that starts at a single top node called the root, like an upside-down family tree. Every node holds a piece of data and can branch out to at most two children below it, known as the left child and the right child. Because there is no ordering rule about which values go left or right, finding a specific item can require checking every single node in the entire tree.
When to reach for it
Reach for binary trees when problems present hierarchical data with left and right child pointers. Questions asking for tree height, maximum depth, path sums from root to leaf, diameter, lowest common ancestor, or checking whether two trees are mirror reflections of each other all signal binary tree traversals. Any problem asking to inspect or reconstruct a tree layer by layer or path by path belongs here.
How the pattern works
Think recursively by focusing on what a single node must do. If the current node is null, return the base answer immediately. Otherwise, ask the left child for its result, ask the right child for its result, and combine both answers with the current node value before returning up to the parent. For horizontal scans, use a queue to read nodes layer by layer, measuring the queue length at the start of each layer to group nodes by depth.
More Binary Trees problems
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