Smallest Subtree with all the Deepest Nodes

Data Structures
Medium
Meta
144.7K views

Given the root of a binary tree, the depth of each node is the shortest distance to the root. Find the node that is the lowest common ancestor of all the deepest nodes in the tree.

Why Interviewers Ask This

Meta interviewers use this problem to assess a candidate's mastery of tree recursion and depth-first search. It specifically tests the ability to handle post-order traversal logic where a node's value depends on its children's return values. The question evaluates if you can efficiently compute depths and identify the Lowest Common Ancestor without redundant traversals, demonstrating strong algorithmic optimization skills.

How to Answer This Question

1. Clarify requirements: Confirm if the tree is balanced or skewed and define 'deepest' as nodes with maximum depth from the root. 2. Propose a recursive strategy: Explain that a single DFS pass can solve this by returning both the maximum depth found in a subtree and the LCA of all deepest nodes within it. 3. Define the base case: If a node is null, return depth zero and null; if it is a leaf, return depth one and itself. 4. Execute the merge logic: Recursively call left and right children. Compare their returned depths. If equal, the current node is the LCA because deepest nodes exist in both subtrees. If unequal, propagate the deeper side up. 5. Optimize complexity: Emphasize that this approach runs in O(N) time and O(H) space, avoiding the brute-force method of calculating depths for every node separately.

Key Points to Cover

  • Identify that a single DFS pass is sufficient rather than multiple traversals
  • Correctly implement post-order logic where parent decisions depend on child results
  • Handle the specific condition where left and right depths are equal to find the LCA
  • Explain the O(N) time complexity advantage over naive approaches
  • Demonstrate clear communication of the recursive state being returned

Sample Answer

To solve the Smallest Subtree with all the Deepest Nodes problem, I would use a post-order depth-first search approach. This allows us to determine the answer in a single pass through the tree. My strategy involves defining a helper function that returns a pair: the maximum depth of the current subtree and the corresponding LCA node. First, I handle the base cases. If the current node is null, I return a depth of zero and a null pointer. If the node is a leaf, the max depth is one, and the node itself is the LCA. Next, I recursively process the left and right children. Let's say the left child returns a depth of dL and the right child returns dR. If dL equals dR, it means the deepest nodes are distributed across both branches at the same level, making the current node the lowest common ancestor. In this scenario, I return the current node with a depth of dL plus one. However, if one side is deeper than the other, the deepest nodes must be entirely within that deeper subtree. Therefore, I simply propagate the result from the deeper child up the tree, incrementing the depth by one. This ensures we always bubble up the correct LCA while tracking the global maximum depth. This solution achieves O(N) time complexity since we visit each node exactly once, and O(H) space for the recursion stack, which aligns well with Meta's expectation for efficient, scalable algorithms.

Common Mistakes to Avoid

  • Calculating depth for every node individually, leading to inefficient O(N^2) performance
  • Forgetting to return both the depth and the node reference from the recursive function
  • Incorrectly handling the case where left and right subtree depths are equal
  • Confusing the depth of the node with the depth of the subtree rooted at that node

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