Minimum Time to Collect All Apples in a Tree

Minimum Time to Collect All Apples in a Tree

What is this problem about?

The Minimum Time to Collect All Apples in a Tree problem presents a tree where some nodes have apples. Starting from the root (node 0), you must visit all nodes containing apples and return to root. Each edge traversal takes 1 unit of time. Find the minimum total time. This Minimum Time to Collect All Apples in a Tree interview question is a tree DFS problem requiring subtree-aware traversal.

Why is this asked in interviews?

Cisco, Microsoft, Meta, Amazon, Google, and Bloomberg ask this because it tests clean DFS implementation on trees — specifically, deciding which subtrees to visit based on whether they contain apples. It validates understanding of post-order traversal and the "only visit if needed" optimization. The BFS, hash table, DFS, and tree interview pattern are all relevant.

Algorithmic pattern used

DFS with subtree pruning. For each node, recursively check if any descendant has an apple. If a subtree contains no apples, don't count its traversal time. If a subtree has at least one apple, add 2 to the time (going down and coming back up the edge to the subtree root). The DFS returns the time needed to collect all apples within a subtree. Sum times across all children that have apples.

Example explanation

Tree: 0-1-2 (path), 0-3 (another child). Apples at nodes 2 and 3.

• DFS(2): leaf with apple → returns 0 (time to collect within subtree). From 1→2: time = 2.
• DFS(1): child 2 has apple, so add 2. Returns 2. From 0→1: time = 2+2 = 4.
• DFS(3): leaf with apple → returns 0. From 0→3: time = 2.
• Total at root: 4 + 2 = 6.

Common mistakes candidates make

• Counting traversal time even for subtrees with no apples.
• Using BFS instead of DFS (DFS naturally computes subtree apple presence).
• Adding 2 for edges to childless (no-apple) subtrees.
• Not building the adjacency list correctly from the edge list input.

Interview preparation tip

Tree traversal problems requiring "visit only needed subtrees" always use DFS with a boolean or integer return value indicating subtree content. The pattern: if any child's DFS returns a positive value (apples found), add 2 for the round trip to that child. Practice this on similar problems like "minimum path to visit all marked nodes" — they all follow the post-order DFS accumulation pattern. Building the adjacency list first, then running DFS from node 0, is the cleanest implementation approach.