btree: avoid duplication of keys

btree/btree.go

@@ -86,7 +86,11 @@ func (t *Tree) Search(k string) (string, error) {
 		return "", fmt.Errorf("The key %s does not exist", k)
 	}
 
-	return t.root.search(&data{key: k})
+	if n, i := t.root.search(&data{key: k}); n != nil {
+		return n.keys[i].value, nil
+	}
+
+	return "", fmt.Errorf("The key %s does not exist", k)
 }
 
 // Remove k in the tree
@@ -123,11 +127,17 @@ func (t *Tree) Insert(key, value string) {
 	// If the tree is empty
 	if t.root == nil {
 		t.root = newNode(t.degree, true)
-		t.root.insertKey(0, k)
+		t.root.writeKey(0, k)
 		t.root.numberOfKeys = 1
 		return
 	}
 
+	// Search for an existing key and replace it
+	if node, i := t.root.search(k); node != nil {
+		node.writeKey(i, k)
+		return
+	}
+
 	// If the tree is not empty
 	if !t.root.isFull() {
 		// If root is not full, insert in non full root
@@ -158,8 +168,8 @@ func (t *Tree) Insert(key, value string) {
 
 // node methods
 
-// insertKey at i place in the node
-func (n *node) insertKey(i int, k *data) {
+// writeKey at i place in the node
+func (n *node) writeKey(i int, k *data) {
 	n.keys[i] = k
 }
 
@@ -192,7 +202,7 @@ func (n *node) visualize(prefix string, isEnd bool) {
 // traverse all nodes in a subtree rooted with this node
 func (n *node) traverse() {
 
-	// There are n entries and n+1 children, treverse trough n keys and n first children
+	// There are n entries and n+1 children, traverse trough n keys and n first children
 	for i := 0; i < n.numberOfKeys; i++ {
 		// If this is not a leaf, then traverse the subtree before printing the key
 		if !n.isLeaf {
@@ -208,7 +218,7 @@ func (n *node) traverse() {
 }
 
 // search k in the subtree rooted with this node
-func (n *node) search(k *data) (string, error) {
+func (n *node) search(k *data) (*node, int) {
 
 	// Find the first entry greater than or equal to k
 	i := 0
@@ -217,16 +227,16 @@ func (n *node) search(k *data) (string, error) {
 	}
 
 	// If the found key is equal to k, return this node
-	if n.keys[i] != nil && k.eq(n.keys[i]) {
-		return n.keys[i].value, nil
+	if i < n.numberOfKeys && k.eq(n.keys[i]) {
+		return n, i
 	}
 
 	// If the key is not found here and this is a leaf node
 	if n.isLeaf {
-		return "", fmt.Errorf("The key %s does not exist", k.key)
+		return nil, -1
 	}
 
-	// Go to the approipriate child
+	// Go to the appropriate child
 	return n.children[i].search(k)
 }
 
@@ -249,24 +259,24 @@ func (n *node) insertNonFull(k *data) {
 		}
 
 		// Insert the new key at the found location
-		n.insertKey(i+1, k)
+		n.writeKey(i+1, k)
 		n.numberOfKeys++
 		return
 	}
 
 	// If this is not a leaf
-	// Finds the child wich is going to have the new key
+	// Finds the child which is going to have the new key
 	for i >= 0 && n.keys[i].gt(k) {
 		i--
 	}
 
-	// Check if the found chird is full
+	// Check if the found child is full
 	if n.children[i+1].isFull() {
 		// If the child is full, then split it
 		n.splitChild(i+1, n.children[i+1])
 
 		// After the split, the middle key of children[i] goes up and
-		// children[i] is splitted into two
+		// children[i] is splited into two
 		// See which of those two is going to have the new key
 		if n.keys[i+1].lt(k) {
 			i++
@@ -352,7 +362,7 @@ func (n *node) remove(k *data) error {
 		isInLastChild = true
 	}
 
-	// If the child where is the key has less than t keys, wi fill it
+	// If the child where is the key has less than t keys, we fill it
 	if n.children[index].numberOfKeys < n.degree {
 		n.fill(index)
 	}

btree/btree_test.go

@@ -84,7 +84,7 @@ func ExampleTree_Traverse() {
 	tree := NewBtree(3)
 
 	toInsert := [][]string{
-		[]string{"a", "A"},
+		[]string{"a", "C"},
 		[]string{"b", "B"},
 		[]string{"c", "C"},
 		[]string{"d", "D"},
@@ -106,6 +106,8 @@ func ExampleTree_Traverse() {
 		tree.Insert(k[0], k[1])
 	}
 
+	tree.Insert("a", "A")
+
 	tree.Visualize()
 	fmt.Println("")