3. Maximum Depth of Binary Tree
easyAsked at RappiCompute the height of a binary tree — Rappi frames this as measuring the longest hand-off chain in a multi-courier relay routing graph.
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Problem
Given the root of a binary tree, return its maximum depth — the number of nodes along the longest path from the root down to the furthest leaf.
Constraints
0 <= nodes <= 10^4-100 <= Node.val <= 100
Examples
Example 1
root = [3,9,20,null,null,15,7]3Example 2
root = [1,null,2]2Approaches
1. Iterative BFS
Level-order traversal incrementing a depth counter per level.
- Time
- O(n)
- Space
- O(n)
let depth = 0, q = root ? [root] : [];
while (q.length) {
const next = [];
for (const n of q) { if (n.left) next.push(n.left); if (n.right) next.push(n.right); }
depth++; q = next;
}
return depth;Tradeoff:
2. Recursive DFS
Depth of a node is 1 plus the max depth of its children; base case empty subtree is 0.
- Time
- O(n)
- Space
- O(h)
function maxDepth(root) {
if (!root) return 0;
return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
}Tradeoff:
Rappi-specific tips
Rappi grades for the recursive solution — they want to see you reason about hand-off chain length the same way their dispatch graph computes worst-case ETA.
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