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Deleting a node from a BST --- Part 1 (easy cases)

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• Warning.....
  • Fact:

    Deleting a node from a Binary Search Tree is very complicated because: The resulting tree after the deletion must also be a Binary Search Tree The best known algorithm used to delete a node from a Binary Search Tree is called: The Hibbard deletion algorithm          Algolist.net: click here

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• Overview: Hibbard delete algorithm
  • Hibbard distinguish 3 cases in the deletion method:

    The deletion node has no subtree (i.e., the deletion node is a leaf node) This case is very easy to handle --- we can simply delete the node The deletion node has only one subtree This case is pretty easy to handle --- we chort-circuit the BST at the deletion node (see example below) The deletion node has two subtrees This case is very tough to handle ....

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• Case 1: deletion node has no subtree (i.e., is a leaf node)
  • Deleting a node that has no subtree (= a leaf node) can be done easily as follows:

    Delete the leaf node by making the link variable in the parent node equal to null Note: Depending on whether the deletion node is the left child or the right child, you must update the left or the right in the parent node !

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    Example 1: if the deletion node is a left child of its parent node

    Delete the node 3 in the following BST: After setting left = null in parent node, we have:

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    Example 2: if the deletion node is a right child of its parent node

    Delete the node 18 in the following BST: After setting right = null in parent node, we have:

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    Special situation: the deletion node is the root node

    If the delete node is the root node.... Note: if the root node has no subtree, then the root node is the only node in the tree: After deleting the root node, we have an empty tree:

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• Case 1: algorithm to delete a node with no subtree
  • Java code used to detect and remove a node with no subtree:

    // p points to the node that you want to remove if ( p.left == null && p.right == null ) // Case 0: p has no subtrees { /* ======================================================= Special case: p is the root node If the root has no subtrees, then it's the ONLY node ! Removing the last node will result in an EMPTY tree ! ======================================================== */ if ( p == root ) { root = null; return; } parent = myParent; // myParent was set by findNode(x).... /* -------------------------------- Delete p from p's parent -------------------------------- */ if ( parent.left == p ) parent.left = null; else parent.right = null; return; }

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• Case 2: deletion node has one subtree
  • Deleting a node that has one subtree can be done easily as follows:

    Delete the node by short-circuiting the BST at the deletion node I.e.: make the parent node of the deletion node point to the subtree of the deletion node The resulting tree is a Binary Search Tree !!!

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    Example 1: if the deletion node only has a left subtree

    Delete the node 5 in the following BST: After linking subtree to the parent node:

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    Example 2: if the deletion node only has a right subtree

    Delete the node 18 in the following BST: After linking subtree to the parent node:

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    Special situation: the deletion node is the root node

    If the root node has only one subtree: We must update the head variable to point to a new root node (which is the subtree of the root node):

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• Case 2: algorithm to delete a node with no subtree
  • Java code used to detect and remove a node with one subtree:

    // p points to the node that you want to remove /* ================================================== Case 1a: p has a right subtree ================================================== */ if ( p.left == null ) // If true, p has a right subtree { /* ======================================== Handle special situation: p is the root ========================================= */ if ( p == root ) { root = root.right; // After root is deleted, BST = right tree of root return; } parent = myParent; // myParent was set by findNode(x).... /* ---------------------------------------------- Link p's RIGHT subtree to the parent node ---------------------------------------------- */ if ( parent.left == p ) { // Hang p's RIGHT tree to parent.left parent.left = p.right; // Link p's right subtree to parent left } else { // Hang p's RIGHT tree to parent.right parent.right = p.right; // Link p's right subtree to parent right } return; } /* ================================================== We must do the same when p has a LEFT subtree Case 1b: p has a left subtree ================================================== */ if ( p.right == null ) // If true, p has a left child { if ( p == root ) { root = root.left; return; } parent = myParent; // myParent was set by findNode(x).... /* ---------------------------------------------- Link p's left child as p's parent child ---------------------------------------------- */ if ( parent.left == p ) parent.left = p.left; else parent.right = p.left; return; }