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Divide and Conquer on Trees

What is this?

Lots of hard-looking tree questions melt away with one habit: solve the small pieces first, then combine. You visit the deepest nodes, get an answer for each little branch, and let those answers bubble back up to their parent — the way you'd add up the cost of every room to get the cost of a whole house. Each spot in the tree returns one number upward while quietly keeping score of the best answer seen so far.

flowchart TD A["Visit a node"] --> B["Solve the left branch"] A --> C["Solve the right branch"] B --> D["Combine both results"] C --> D D --> E["Return one value to the parent"]

💡 Fun fact: This bottom-up combining is a baby version of dynamic programming — you compute each subproblem exactly once, so even a giant tree is solved in a single sweep.

🔓 The 6 problems in this chapter are free. Sign in with Google or Microsoft to start solving.


Core idea: A huge family of tree problems is solved by one move: a post-order DFS where each call returns a single value to its parent, while updating a shared (global/nonlocal) answer that may combine both children. The return value is what a parent can extend; the answer is what bends at the current node. Spotting which information flows up (returned) versus down (passed as arguments) is the whole skill.


The pattern

The dfs template: recurse on both children, update the global answer with combine(L, R, node) using both branches (the bend), and return contribution_to_parent(L, R, node), one branch upward

The recurring tension: a parent can only continue one branch through a child, but the optimal answer often uses both branches meeting at a node. So you return one, but score with both.


The problems

Problem family map: DnC on trees splits into bottom-up problems (Diameter, Max Path Sum, Longest Univalue Path, Smallest Subtree of Deepest Nodes, LCA II) and top-down (Max Diff Node-Ancestor)

  • Diameter, Max Path Sum, Longest Univalue Path — the canonical trio: return an arm/height/gain upward, update the answer with left + right.
  • Smallest Subtree with all the Deepest Nodes — return a richer tuple (depth, covering-node); equal child depths mean this node is the answer.
  • LCA II — fuse an existence check into the search with a found-counter, trusting the result only if both targets were truly seen.
  • Maximum Difference Between Node and Ancestor — the mirror image: information flows down (carry the path min/max as arguments).

Key takeaways

  • Post-order DFS: return one thing, update another — the engine behind diameter, path sum, and more.
  • Return one branch upward; the answer may bend using both branches at a node.
  • Richer return types (tuples, counters) collapse multi-pass solutions into one O(n) pass.
  • Direction matters: bottom-up returns info; top-down carries info down as arguments.
  • Why interviewers love it: it tests whether you can design what each recursive call communicates.

Order: Diameter → Maximum Path Sum → Longest Univalue Path → LCA II → Smallest Subtree of Deepest Nodes → Max Difference Node–Ancestor.

Smallest Subtree with all the Deepest Nodes

Core idea: The answer is the lowest common ancestor (LCA) of all the deepest nodes — the deepest point from which you can still "see" every maximum-depth node below you. You don't need two passes to find it. A single post-order DFS that returns a tuple (depth, covering-node) for each subtree computes depth and the answer together: if a node's two children report equal depth, the deepest nodes live on both sides, so this node is their meeting point; if one side is strictly deeper, the answer is whatever that deeper side already reported.


Problem, rephrased

Forget the textbook phrasing. Here's the scenario:

You manage an org chart rooted at the CEO. Some employees sit further down the reporting chain than others — measure each person's depth as the number of edges from the CEO. The people at the maximum depth are the "frontline" — the furthest-removed reports. You want to name the single lowest manager whose org sub-tree still contains every frontline person. Equivalently: return the root of the smallest subtree that contains all of the tree's deepest (maximum-depth) nodes.

You're handed the root of a binary tree. Return the node that roots that smallest covering subtree.

Take this little tree (depths annotated on the right):

Chalkboard sketch of the tree rooted at 3 with depth labels 0 to 3; the deepest nodes at depth 3 are 7 and 4

The deepest nodes Their smallest covering subtree (the answer) Why
7 and 4 (depth 3) node 2 2 is the lowest node that has both 7 and 4 underneath it — their LCA

If there were only one deepest node, the answer would simply be that node itself.


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