Left View of a Binary Tree

Given the root of a binary tree, return its left view as a list of node values from top to bottom.
At each depth, include the leftmost existing node when the nodes at that depth are ordered from left to right. Include one value for every nonempty depth, even when values repeat.
An empty tree returns an empty list. The tree is displayed in level order, using null for missing children.

Function
leftView(root: TreeNode) → List<Integer>

Examples
Example 1
root = [1,2,3,null,4,5,6]
return = [1,2,4]
The first nodes at depths 0, 1, and 2 have values 1, 2, and 4.
Example 2
root = [8,null,9,null,10]
return = [8,9,10]
There is one existing node at each depth. All three are visible despite having only right-child links.

Constraints
The tree has 0 to 5000 nodes.
Each node value is an integer in [-10^6, 10^6].
Each node has at most two children, and the input is a valid tree with no cycles or shared child nodes.

The trick is that left view means first node per depth, not following left child pointers. Two clean ways. BFS: process level by level, and record the first node dequeued at each level. DFS: visit the node, then left, then right, passing depth, and add the value only when depth equals the current result size. Both are O(n) time. The classic pitfall is walking only left children, which fails example 2 where every child is a right child. Another is handling the empty tree, which must return an empty list. With up to 5000 nodes, recursion depth could reach 5000 on a skewed tree, so BFS is the safer pick in some languages. Duplicate values don't matter, since you key on depth, not value. If you blank mid-assessment, StealthCoder is the hedge that reads the problem and hands you the level-order version.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.