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Binary Search Trees. Adding and Removing. Adding to the Tree. Adding to a binary search tree is fairly straightforward. You will begin at the root, and as you move down to each node, there are three cases: 1. The value to be added is equal to the value of the current node.
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Binary Search Trees Adding and Removing
Adding to the Tree Adding to a binary search tree is fairly straightforward. You will begin at the root, and as you move down to each node, there are three cases: 1. The value to be added is equal to the value of the current node. 2. The value to be added is less than the value of the current node. 3. The value to be added is greater than the value of the current node. Lesser values should always be added to the left subtree while greater values should always be added to the right subtree. A value equal to the current node should never be added to the tree. The tree below was created by adding values 10, 5, 15, 13, 1, 17, 7, 6, 2, and 9 in that order.
Removing from the Tree Removing from a binary search tree is much more difficult. There are many logically correct ways to remove from a tree, but only one correct way to pass the test driver. Carefully consider the following cases when writing your remove method.
Case 1: The Leaf Node The node to be removed is a leaf node. A leaf node has no children. Nodes 2, 6, 9, 13, and 17 are all leaf nodes because they have neither a left child nor a right child. In the example below, the leaf node 2 is removed. Node 1’s right child is assigned to NULL. Before After
Case 2: The Inorder Predecessor The inorder predecessor is the rightmost node in the left subtree of the node to be removed. The node to be removed should be replaced with its inorder predecessor. This allows us to remove nodes with children without rebuilding the entire tree. The node to be removed (node 10) is not a leaf, so there is more work to do. The inorder predecessor in this case is node 9, so we set node 10’s value to 9 and remove node 9 easily because it is a leaf. Before After
Case 3: The Inorder Predecessor with a Left Child An inorder predecessor will never have a right child, but it might have a left child. In this case, the predecessor’s parent adopts the predecessor’s left child as its right child. In this case, we are removing node 9. The predecessor is node 7, but 7 has a child. When we remove node 7, we must reassign node 5’s right child to point to node 6. Before After
Case 4: The Predecessor is the Left Child When finding the inorder predecessor, you first move to the left child of the node to be removed, then move as far right as possible. If that left child does not have a right child, then it is the inorder predecessor. Like before, the node to be removed takes the predecessor’s value. The node to be removed then adopts the predecessor’s left child as its own left child. In this example, we want to remove node 15. We look to node 13 to find the inorder predecessor; node 13 has no right child. In this case, we assign node 15 a value of 13 and set node 15’s left child to the child of 13, NULL in this case. Before After
Case 5: The Missing Predecessor If the node to be removed does not have a left child, there is no inorder predecessor. In this case, the node to be removed takes the value of its right child and then adopts its right child’s children as its own children. In this case, we want to remove node 13, which has no left child. We assign node 13 a value of 17 and set 13’s left child to node 16 and its right child to node 18. Before After
