trie: small optimization of delete in fullNode case (#22979)
When deleting in fullNode, and the new child node nn is not nil, there is no need to check the number of non-nil entries in the node. This is because the fullNode must've contained at least two children before deletion, so there must be another child node other than nn. Co-authored-by: Felix Lange <fjl@twurst.com>
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@ -405,6 +405,14 @@ func (t *Trie) delete(n node, prefix, key []byte) (bool, node, error) {
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n.flags = t.newFlag()
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n.Children[key[0]] = nn
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// Because n is a full node, it must've contained at least two children
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// before the delete operation. If the new child value is non-nil, n still
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// has at least two children after the deletion, and cannot be reduced to
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// a short node.
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if nn != nil {
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return true, n, nil
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}
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// Reduction:
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// Check how many non-nil entries are left after deleting and
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// reduce the full node to a short node if only one entry is
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// left. Since n must've contained at least two children
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