PPractick

The problem

A photographer shoots a tree sculpture from the right side and then from the left side. From the right, each level shows its rightmost node; from the left, its leftmost. A DFS that tracks depth captures either view in one pass. For the right view, visit the right child before the left; the first node reached at each new depth is the visible one. For the left view, visit left first.

Example: [1, 2, 3, null, 5, null, 4] → right view [1, 3, 4], left view [1, 2, 5].

Function signature

vector<int> rightSideView(TreeNode* root);
vector<int> leftSideView(TreeNode* root);

Test cases

Input (level order)	Output	Explanation
[1, 2, 3, null, 5, null, 4]	right [1, 3, 4] left [1, 2, 5]
[1, 2, 3, 4]	right [1, 3, 4] left [1, 2, 4]	4 shows from both sides
[1, null, 3]	right [1, 3] left [1, 3]
[1, 2]	right [1, 2] left [1, 2]
[]	right [] left []

(Confirm: one node per level, top to bottom; a node hidden behind others on its level isn't visible.)

How interviewers phrase it

• "Imagine yourself on the right side of a binary tree; return the values of the nodes you can see, ordered from top to bottom" (LeetCode 199), with the left view as a follow-up (GeeksforGeeks "Left View of Binary Tree").

Clarifying questions to ask

1. Top to bottom order? (Yes.)
2. Is the rightmost node of each level visible even if it's in the left subtree? (Yes, as with 5 in the first example's left view, or 4 in the second example's right view.)
3. Can the tree be empty?

Follow-ups to expect

1. "DFS approach?"
2. "BFS approach?"
3. "Why does the visit order matter for DFS?"
4. "How does this relate to top and bottom views?"
5. "Complexity?"

Related questions

• Vertical Order Traversal Medium
• Print all nodes at a distance of K in BT Hard
• Minimum time taken to burn the BT from a given Node Hard
• Maximum Depth in BT Medium
• Check for balanced binary tree Medium
• Maximum path sum Medium

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