/*
* case 1. item == null, not found.
* case 2: item < node.data, search left.
* case 3. item > node.data, search right.
* case 4. item not in tree, do nothing.
* case 5: item = node.data, remove it.
* (a) if node has a single child, attach the child
* to node's parent
* (b) if node has two children,
* replace node with the largest node in its left sub tree
*/
private BSTNode remove(BSTNode node, int data) {
if (node == null) {
return node;
}
if (data < node.data) {
node.left = remove(node.left, data);
return node;
}
if (data > node.data) {
node.right = remove(node.right, data);
return node;
} else {
// Once item found, remove
// If has single child, replace parent with child
if (node.left == null) {
return node.right;
} else if (node.right == null) {
return node.left;
}
// If has two children, replace the largest node on left
else {
node.data = findMax(node.left);
return node;
}
}
} // End remove(Node)
/**
* Return the largest child of the node parent which is the right most node in the left sub tree.
*/
private int findMax(BSTNode parent) {
if (parent.right.right == null) {
int value = parent.right.data;
parent.right = parent.right.left;
return value;
} else {
return findMax(parent.right);
}
} // End findLargestChild(Node))
private BSTNode insert(BSTNode node, int data) {
if (node == null) {
node = new BSTNode(data);
} else {
// This allows duplicates, however
if (data <= node.data) {
node.left = insert(node.left, data);
} else {
node.right = insert(node.right, data);
}
}
return node;
}