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 β 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.
π‘ Fun fact: This bottom-up combining is a baby version of dynamic programming β you compute each subproblem exactly once, so even a giant tree is solved in a single sweep.
π 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 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
- Diameter, Max Path Sum, Longest Univalue Path β the canonical trio: return an arm/height/gain upward, update the answer with
left + right. - Smallest Subtree with all the Deepest Nodes β return a richer tuple
(depth, covering-node); equal child depths mean this node is the answer. - LCA II β 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 β the mirror image: information flows down (carry the path min/max as arguments).
Key takeaways
- Post-order DFS: return one thing, update another β 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 β Maximum Path Sum β Longest Univalue Path β LCA II β Smallest Subtree of Deepest Nodes β Max Difference NodeβAncestor.
Smallest Subtree with all the Deepest Nodes
Core idea: The answer is the lowest common ancestor (LCA) of all the deepest nodes β 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 β measure each person's depth as the number of edges from the CEO. The people at the maximum depth are the "frontline" β 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):
| 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 β their LCA |
If there were only one deepest node, the answer would simply be that node itself.
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