Question

In a red-black tree the following vertices are found on a path from the root to a leaf: black, red, black, black. What is the minimum possible number of keys stored in the tree?

Answer 1

Minimum possible number of keys stored in the tree is 5.

Lets look an eg.

Answer 2

Calculation:

1. Path Structure: The given path has 4 nodes in the order: Black (B), Red (R), Black (B), Black (B).

• Let’s denote the path as: B₁ → R₂ → B₃ → B₄.

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2. Red-Black Tree Rules:

• Black Root: B₁ is the root and must be black.

• No Double Reds: A red node (R₂) cannot have a red parent or child.

• Equal Black Nodes: Every path from the root to a leaf must have the same number of black nodes (Black Height = 3 for this path).

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3. Minimizing Keys:

• B₁: Must have at least 1 key (as the root).

• R₂: Must have 2 children (since it’s red and cannot have a red child, its children must be black). Assign the minimum keys (1 key per child).

• B₃: On the given path, it’s part of the sequence. Its other child (not on the path) must be a leaf (Nil) to minimize keys. Assign 1 key to B₃.

• B₄: Must be a leaf (no keys, but counts as a black node).

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4. Other Subtrees:

• The left subtree of B₁ must have a black height of 2 (to match the right subtree’s black height of 3 minus 1). The minimal structure is a black root with two red children (each with two Nil leaves).

• Total keys in the left subtree: 3 (B → R → R).

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5. Total Keys:

• Left subtree of B₁: 3 keys.

• B₁: 1 key.

• Right subtree (B₁ → R₂ → B₃ → B₄): R₂ has 1 key, B₃ has 1 key, and the other child of R₂ is a black leaf (1 key).
  Total in right subtree: 1 (R₂) + 1 (B₃) + 1 (other child) = 3 keys.

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• Sum: 3 (left) + 1 (root) + 3 (right) + 3 (other subtrees under R₂) = 10 keys.

Thus, the minimum number of keys is 10