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DSA Javascriptprogramming~5 mins

BST Delete Operation in DSA Javascript - Cheat Sheet & Quick Revision

Choose your learning style10 modes available
Recall & Review
beginner
What are the three main cases to consider when deleting a node in a Binary Search Tree (BST)?
1. Node to delete is a leaf (no children).
2. Node to delete has one child.
3. Node to delete has two children.
Click to reveal answer
intermediate
In BST deletion, how do you handle deleting a node with two children?
Replace the node's value with its inorder successor (smallest value in right subtree) or inorder predecessor (largest value in left subtree), then delete that successor/predecessor node.
Click to reveal answer
beginner
What is the inorder successor in a BST?
The node with the smallest value in the right subtree of the current node.
Click to reveal answer
intermediate
Why do we replace a node with its inorder successor or predecessor when deleting a node with two children?
Because it maintains the BST property by ensuring all left subtree nodes are smaller and all right subtree nodes are larger than the replaced node.
Click to reveal answer
beginner
What happens if you delete a leaf node in a BST?
You simply remove the leaf node by setting its parent's pointer to null.
Click to reveal answer
Which of the following is NOT a case to consider when deleting a node in a BST?
ANode has no children
BNode has one child
CNode has three children
DNode has two children
When deleting a node with two children, which node is commonly used to replace it?
AInorder successor
BRoot node
CRandom leaf node
DParent node
What is the inorder successor of a node in a BST?
ASmallest node in right subtree
BParent node
CLargest node in left subtree
DAny leaf node
If a node to delete has only one child, what is the deletion process?
AReplace node with inorder successor
BRemove the node and connect its child directly to its parent
CReplace node with inorder predecessor
DDo nothing
What happens to the BST property after deleting a node and replacing it with its inorder successor?
ABST becomes empty
BBST property is broken
CBST becomes a linked list
DBST property is maintained
Explain the steps to delete a node with two children in a BST.
Think about how to keep the tree ordered after deletion.
You got /4 concepts.
    Describe how deleting a leaf node differs from deleting a node with one child in a BST.
    Consider the connections between nodes.
    You got /4 concepts.

      Practice

      (1/5)
      1. What happens when you delete a leaf node (a node with no children) from a Binary Search Tree (BST)?
      easy
      A. The node is simply removed without affecting other nodes.
      B. The entire tree is rebuilt from scratch.
      C. The node's children replace it in the tree.
      D. The root node is always changed.

      Solution

      1. Step 1: Understand leaf node deletion

        A leaf node has no children, so removing it does not affect other nodes.
      2. Step 2: Effect on BST structure

        Since no children exist, simply remove the node and update its parent's pointer to null.
      3. Final Answer:

        The node is simply removed without affecting other nodes. -> Option A
      4. Quick Check:

        Leaf node deletion = simple removal [OK]
      Hint: Leaf nodes delete by direct removal only [OK]
      Common Mistakes:
      • Thinking children replace the node when none exist
      • Assuming root always changes
      • Believing tree rebuild is needed
      2. Which of the following is the correct syntax to delete a node with value key from a BST in JavaScript?
      easy
      A. bst.deleteNode(key);
      B. bst.delete(key);
      C. bst.remove(key);
      D. bst.erase(key);

      Solution

      1. Step 1: Identify common BST delete method

        In many JavaScript BST implementations, the method to delete a node is named delete.
      2. Step 2: Check method naming conventions

        Methods like remove, erase, or deleteNode are not standard or common in JavaScript BST examples.
      3. Final Answer:

        bst.delete(key); -> Option B
      4. Quick Check:

        Standard BST delete method = delete() [OK]
      Hint: Use bst.delete(key) to remove nodes [OK]
      Common Mistakes:
      • Using remove() or erase() which are not standard
      • Confusing delete keyword with method name
      • Using deleteNode() which is uncommon
      3. Consider this BST and the delete operation:
          10
         /  \
        5    15
             / \
            12  20
      
      What will be the in-order traversal output after deleting node with value 15?
      medium
      A. [5, 10, 12, 15, 20]
      B. [5, 10, 20]
      C. [5, 10, 12, 20]
      D. [5, 10, 15, 20]

      Solution

      1. Step 1: Identify node to delete and its children

        Node 15 has two children: 12 (left) and 20 (right).
      2. Step 2: Replace node 15 with its in-order successor

        The in-order successor is the smallest node in right subtree, which is 20. Replace 15 with 20 and delete original 20 node.
      3. Step 3: Perform in-order traversal

        In-order traversal visits nodes in ascending order: 5, 10, 12, 20.
      4. Final Answer:

        [5, 10, 12, 20] -> Option C
      5. Quick Check:

        In-order after deleting 15 = [5, 10, 12, 20] [OK]
      Hint: Replace deleted node with in-order successor [OK]
      Common Mistakes:
      • Removing 15 without replacement
      • Using in-order predecessor instead of successor
      • Forgetting to delete the successor node
      4. The following code attempts to delete a node from a BST but causes an error. What is the main issue?
      function deleteNode(root, key) {
        if (!root) return null;
        if (key < root.val) root.left = deleteNode(root.left, key);
        else if (key > root.val) root.right = deleteNode(root.right, key);
        else {
          if (!root.left) return root.right;
          else if (!root.right) return root.left;
          let minNode = findMin(root.right);
          root.val = minNode.val;
          // Missing deletion of minNode here
        }
        return root;
      }
      
      function findMin(node) {
        while (node.left) node = node.left;
        return node;
      }
      medium
      A. The function does not handle leaf nodes correctly.
      B. The base case for recursion is missing.
      C. The function uses incorrect comparison operators.
      D. The code does not delete the in-order successor node after copying its value.

      Solution

      1. Step 1: Analyze deletion of node with two children

        When deleting a node with two children, we replace its value with the in-order successor's value.
      2. Step 2: Check if in-order successor node is deleted

        The code copies the successor's value but does not delete the successor node itself, causing duplicate values and incorrect tree.
      3. Final Answer:

        The code does not delete the in-order successor node after copying its value. -> Option D
      4. Quick Check:

        Must delete successor node after value copy [OK]
      Hint: Always delete successor node after copying value [OK]
      Common Mistakes:
      • Forgetting to delete the in-order successor node
      • Missing base case (though present here)
      • Incorrectly handling leaf nodes
      • Using wrong comparison operators
      5. You have a BST with nodes: 8, 3, 10, 1, 6, 14, 4, 7, 13.
      After deleting node 3, what is the correct in-order traversal of the BST?
      hard
      A. [1, 4, 6, 7, 8, 10, 13, 14]
      B. [1, 3, 4, 6, 7, 8, 10, 13, 14]
      C. [1, 6, 7, 8, 10, 13, 14]
      D. [1, 4, 7, 8, 10, 13, 14]

      Solution

      1. Step 1: Identify node 3 and its children

        Node 3 has two children: 1 (left) and 6 (right). 6 has children 4 and 7.
      2. Step 2: Delete node 3 by replacing with in-order successor

        The in-order successor of 3 is 4 (smallest in right subtree). Replace 3 with 4, then delete original 4 node.
      3. Step 3: Perform in-order traversal after deletion

        In-order traversal visits nodes in ascending order: 1, 4, 6, 7, 8, 10, 13, 14.
      4. Final Answer:

        [1, 4, 6, 7, 8, 10, 13, 14] -> Option A
      5. Quick Check:

        In-order after deleting 3 = [1, 4, 6, 7, 8, 10, 13, 14] [OK]
      Hint: Replace deleted node with in-order successor, then traverse [OK]
      Common Mistakes:
      • Forgetting to delete the in-order successor node
      • Using in-order predecessor instead
      • Skipping nodes in traversal
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