Bottom View of a Binary Tree - Iterative Approach

Contents

ProblemExamplesVisualizationSolutionAlgorithmCode

Problem Statement❯

Given a binary tree, return the bottom view of the tree. The bottom view consists of the nodes that are visible when the tree is viewed from the bottom. For each horizontal distance from the root, the node that is at the lowest level (deepest) will be visible. Your task is to determine these visible nodes using an iterative approach.

Examples❯

Input Tree Bottom View Output Description
[20, 8, 22, 5, 3, null, 25, null, null, 10, 14]
[5, 10, 3, 14, 25] Standard binary tree showing nodes visible from the bottom across horizontal distances
[1]
[1] Single node tree; only the root is visible from the bottom
[] [] Empty tree; no nodes to display in bottom view
[1, 2, null, 3, null, 4]
[4, 2, 1] Left-skewed tree; last node at each horizontal distance is shown
[1, null, 2, null, null, null, 3]
[1, 2, 3] Right-skewed tree; all nodes fall on unique horizontal distances
[1, 2, 3, 4, 5, 6, 7]
[4, 2, 6, 3, 7] Complete binary tree; bottom-most nodes overwrite top ones at each horizontal distance

Visualization Player

Solution❯

Case 1: General Binary Tree

In a general binary tree where nodes exist on both left and right sides of the root, the bottom view is determined by traversing the tree level by level (breadth-first). At each horizontal distance from the root, we keep updating the value in a map until we reach the last node at that distance. For example, in the first example tree, node 3 and node 10 share the same horizontal distance, but 10 appears later in the level order traversal, and it's deeper, so it takes the bottom view spot.

Case 2: Right-Skewed Tree

In a right-skewed binary tree, each node appears one level deeper than the previous and shifts the horizontal distance by +1 each time. This ensures all nodes appear in the bottom view since no node hides any other due to unique horizontal distances.

Case 3: Left-Skewed Tree

Similarly, in a left-skewed binary tree, each node is added one level deeper and to the left of the previous node (horizontal distance -1 each time). Hence, every node gets its own position in the bottom view when seen from below.

Case 4: Empty Tree

If the input tree is empty, then there's nothing to show in the bottom view. The result will be an empty list because there are no nodes to process.

How it works

We traverse the tree using a queue (level order), and along with each node, we track its horizontal distance (hd) from the root. At every step, we update a map with the latest node at that hd — meaning the last one visited at that distance, which will be the bottommost. After the traversal, sorting the keys of this map gives the left-to-right bottom view.

Algorithm Steps❯

  1. If the binary tree is empty, return an empty result.
  2. Initialize a queue and enqueue the root node along with its horizontal distance (hd = 0).
  3. Initialize an empty map (hdMap) to store the latest node value at each horizontal distance.
  4. While the queue is not empty, dequeue an element (node, hd).
  5. Update hdMap[hd] with the node's value (this ensures the bottom-most node at that hd is recorded).
  6. If the node has a left child, enqueue it with hd - 1; if it has a right child, enqueue it with hd + 1.
  7. After processing all nodes, sort the keys of hdMap and record their corresponding values. The ordered values form the bottom view of the binary tree.

Code

Python
Java
JavaScript
C
C++
C#
Go
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

from collections import deque

def bottomView(root):
    if not root:
        return []
    q = deque([(root, 0)])
    hdMap = {}
    while q:
        node, hd = q.popleft()
        hdMap[hd] = node.val
        if node.left:
            q.append((node.left, hd - 1))
        if node.right:
            q.append((node.right, hd + 1))
    result = [hdMap[hd] for hd in sorted(hdMap)]
    return result

if __name__ == '__main__':
    # Construct binary tree:
    #         20
    #        /  \
    #       8    22
    #      / \     \
    #     5   3     25
    #        / \
    #       10  14
    root = TreeNode(20,
                    TreeNode(8, TreeNode(5), TreeNode(3, TreeNode(10), TreeNode(14))),
                    TreeNode(22, None, TreeNode(25)))
    print(bottomView(root))