Lowest Common Ancestor of a Binary Tree IV

Lowest Common Ancestor of a Binary Tree IV

What is this problem about?

The Lowest Common Ancestor of a Binary Tree IV is an expansion of the classic LCA problem. Instead of being given just two nodes (p and q), you are given a generic binary tree and an array containing multiple target nodes. Your task is to find the lowest common ancestor for all the nodes in this entire array.

Why is this asked in interviews?

Interviewers use this problem to see if you can scale a fundamental algorithm to handle collections of data. It tests whether you can adapt the standard post-order DFS LCA template to process a Hash Set of targets. It evaluates your ability to combine Tree Traversal with Hash Set lookups to maintain efficient $O(N)$ time complexity instead of running multiple pairwise LCA searches.

Algorithmic pattern used

This problem relies on a modified Post-Order Depth-First Search (DFS) combined with a Hash Set. First, you dump all target nodes into a Hash Set for $O(1)$ lookups. Then, you traverse the tree. If the current node exists in the Set, you immediately return it. Otherwise, you recursively search the left and right subtrees. If both subtrees return a non-null value (meaning targets were found in both branches), the current node is the LCA for the targets below it.

Example explanation

Imagine a tree with root 3, left 5, right 1. Under 5 are 6 and 2. Target array: [6, 2, 1]. Convert array to Set: {6, 2, 1}. DFS from 3:

• DFS Left to 5.
  • DFS Left to 6. It's in the Set! Return 6.
  • DFS Right to 2. It's in the Set! Return 2.
  • At node 5, both left (6) and right (2) are non-null. Node 5 is their LCA. Return 5 up to the root.
• DFS Right to 1. It's in the Set! Return 1 up to the root.
• At root 3, the left returned 5 (which covers targets 6 and 2) and the right returned 1 (which covers target 1). Since both sides returned non-null, root 3 is the overall LCA.

Common mistakes candidates make

A frequent mistake is solving this by iterating through the array and finding the LCA of the first two nodes, then finding the LCA of that result with the third node, and so on. While this works mathematically, it results in $O(K \times N)$ time complexity, where $K$ is the number of nodes. The Hash Set + single DFS approach is strictly $O(N)$.

Interview preparation tip

For the Lowest Common Ancestor IV coding problem, remember that the core LCA logic doesn't change based on the number of nodes! The only difference is checking if (set.contains(root)) instead of if (root == p || root == q). Keeping your base cases dynamic prevents unnecessary code bloat.