PPractick

The problem

Now look at the tree from below. For each horizontal distance (root 0, left −1, right +1), the visible node is the lowest one. If two nodes share the lowest level at the same distance, the one appearing later in level order wins. A BFS that keeps overwriting the entry for each distance ends with exactly that node, because BFS visits deeper levels later and, within a level, from left to right.

Example: [20, 8, 22, 5, 3, null, 25, null, null, 10, 14] → [5, 10, 3, 14, 25].

Function signature

vector<int> bottomView(TreeNode* root);

Test cases

Input (level order)	Output	Explanation
[20, 8, 22, 5, 3, null, 25, null, null, 10, 14]	[5, 10, 3, 14, 25]	10 and 14 hide 8 and 22
[20, 8, 22, 5, 3, 4, 25, null, null, 10, 14]	[5, 10, 4, 14, 25]	3 and 4 tie at distance 0; the later one (4) wins
[1, 2, 3]	[2, 1, 3]
[1]	[1]
[]	[]

(Confirm: the lowest node per horizontal distance; ties on the same level go to the one later in level order.)

How interviewers phrase it

• "Print the bottom view of a binary tree: for each horizontal distance, the bottom-most node; if there are several, the later one in level order" (GeeksforGeeks "Bottom View of Binary Tree").

Clarifying questions to ask

1. How are ties at the same depth and distance broken? (The later one in level order.)
2. How is horizontal distance defined?
3. Left-to-right output? (Yes, by increasing distance.)

Follow-ups to expect

1. "How does this differ from the top view?"
2. "Why is the last node seen the correct one?"
3. "Would DFS work?"
4. "Left or right views?"
5. "Complexity?"

Related questions

• Vertical Order Traversal Medium
• Top View of BT Medium
• Zig Zag or Spiral Traversal Medium
• Right/Left View of Binary Tree Medium
• Maximum Width of BT Medium
• Level Order Traversal Easy

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