Top View of a Binary Tree - Iterative Approach

Problem Statement❯

Given the root of a binary tree, return the top view of the tree. The top view of a binary tree is the set of nodes visible when the tree is viewed from the top. You should return the values of these nodes from left to right, based on their horizontal distance from the root. Use an iterative approach (like level-order traversal) to compute this top view.

Examples❯

Input Tree	Top View Output	Description
[1, 2, 3, 4, 5, null, 6]	[4, 2, 1, 3, 6]	Standard tree with multiple levels; leftmost and rightmost nodes at each level appear in top view
[1]	[1]	Single node tree; root itself is the top view
[]	[]	Empty tree with no nodes returns an empty top view
[1, 2, null, 3, null, 4]	[4, 3, 2, 1]	Left-skewed tree; only the leftmost path visible in top view
[1, null, 2, null, 3, null, 4]	[1, 2, 3, 4]	Right-skewed tree; every node appears in top view due to unique horizontal distances
[1, 2, 3, 4, null, null, 5, 6]	[6, 4, 2, 1, 3, 5]	Complex tree with deep left and right subtrees; top view includes the outermost nodes

Visualization Player

Solution❯

Case 1: Empty Tree

If the binary tree is empty (i.e., the root is null), then there are no nodes to observe from the top. Therefore, the top view should be an empty list.

Case 2: Single Node Tree

If the tree contains only a single root node, then that node is visible from the top view as there are no other nodes to obstruct it.

Case 3: Left-Skewed Tree

In a left-skewed tree, all nodes only have left children. As we go down the tree, each node shifts one unit to the left in horizontal distance. All of them will be visible from the top as no node shares the same horizontal distance.

Case 4: Right-Skewed Tree

This is the opposite of the left-skewed tree. All nodes have only right children. Each node moves one step to the right in horizontal distance and remains visible from the top because there is no overlap in horizontal levels.

Case 5: General Tree with Both Children

In a typical binary tree with both left and right children, we perform a level-order traversal (BFS) while tracking the horizontal distance (HD) from the root. The root is at HD = 0, left child HD = -1, right child HD = +1, and so on. For the top view, we want the first node encountered at each HD. So, we use a map to store the node value at each HD the first time it's visited. This ensures we capture only the uppermost node at every horizontal position.

Once the traversal completes, we sort the horizontal distances and return the corresponding values to get the left-to-right top view of the tree.

Algorithm Steps❯

If the tree is empty, return an empty result.
Initialize a queue and enqueue a tuple containing the root node and its horizontal distance (0).
Initialize an empty map (or dictionary) to store the first node encountered at each horizontal distance.
While the queue is not empty, dequeue a tuple (node, hd).
If the horizontal distance hd is not present in the map, record the node's value for that hd.
Enqueue the left child with horizontal distance hd - 1 and the right child with hd + 1 (if they exist).
After processing all nodes, sort the keys of the map and output the corresponding node values as the top view.

Code

Python

Java

JavaScript

C++

Kotlin

Swift

Php

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def topView(root):
    if not root:
        return []
    from collections import deque
    queue = deque([(root, 0)])
    hd_map = {}
    while queue:
        node, hd = queue.popleft()
        if hd not in hd_map:
            hd_map[hd] = node.val
        if node.left:
            queue.append((node.left, hd - 1))
        if node.right:
            queue.append((node.right, hd + 1))
    return [hd_map[hd] for hd in sorted(hd_map)]

# Example usage:
if __name__ == '__main__':
    # Construct binary tree:
    #         1
    #        / \
    #       2   3
    #      / \   \
    #     4   5   6
    root = TreeNode(1, TreeNode(2, TreeNode(4), TreeNode(5)), TreeNode(3, None, TreeNode(6)))
    print(topView(root))

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