BST Inorder Successor in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the minimum node in a BST subtree.

DSA Go

func minValueNode(node *Node) *Node {
    current := node
    for current.[1] != nil {
        current = current.left
    }
    return current
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' will find the maximum, not minimum.

Using 'parent' or 'next' is incorrect as they are not child pointers.

The minimum value node in a BST subtree is found by going as left as possible.

2fill in blank

medium

Complete the code to find the inorder successor when the node has a right child.

DSA Go

func inorderSuccessor(root, n *Node) *Node {
    if n.right != nil {
        return [1](n.right)
    }
    // other cases omitted
    return nil
}

Drag options to blanks, or click blank then click option'

AminValueNode

BmaxValueNode

CfindParent

DinorderPredecessor

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxValueNode will give the largest node, not successor.

Using inorderPredecessor is the opposite of successor.

If the node has a right child, the successor is the minimum node in the right subtree.

3fill in blank

hard

Fix the error in the loop to find the inorder successor when the node has no right child.

DSA Go

func inorderSuccessor(root, n *Node) *Node {
    var successor *Node
    current := root
    for current != nil {
        if n.val < current.val {
            successor = current
            current = current.[1]
        } else {
            current = current.right
        }
    }
    return successor
}

Drag options to blanks, or click blank then click option'

Anext

Bright

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Moving right when node value is less will skip potential successors.

Using parent or next pointers is not applicable here.

When searching for successor without right child, move left if node value is less than current.

4fill in blank

hard

Fill both blanks to complete the inorder successor function for all cases.

DSA Go

func inorderSuccessor(root, n *Node) *Node {
    if n.right != nil {
        return [1](n.right)
    }
    var successor *Node
    current := root
    for current != nil {
        if n.val < current.val {
            successor = current
            current = current.[2]
        } else {
            current = current.right
        }
    }
    return successor
}

Drag options to blanks, or click blank then click option'

AminValueNode

BmaxValueNode

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxValueNode or moving right in the loop causes wrong successor.

Not handling both cases leads to incorrect results.

Use minValueNode for right subtree and move left in the loop to find successor without right child.

5fill in blank

hard

Fill all three blanks to complete a function that returns the inorder successor value or -1 if none.

DSA Go

func inorderSuccessorValue(root, n *Node) int {
    successor := inorderSuccessor(root, n)
    if successor != nil {
        return successor.[1]
    }
    return [2]
}

func inorderSuccessor(root, n *Node) *Node {
    if n.right != nil {
        return [3](n.right)
    }
    var successor *Node
    current := root
    for current != nil {
        if n.val < current.val {
            successor = current
            current = current.left
        } else {
            current = current.right
        }
    }
    return successor
}

Drag options to blanks, or click blank then click option'

Aval

B-1

CminValueNode

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the node pointer instead of value.

Returning 0 or other default instead of -1.

Using wrong function for right subtree.

Return the value of successor node, -1 if none, and use minValueNode for right subtree.

Practice

1. What is the inorder successor of a node in a Binary Search Tree (BST)?

easy

A. The node with the largest key smaller than the given node's key

B. The node with the smallest key greater than the given node's key

C. The root node of the BST

D. The left child of the given node

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits nodes in ascending order of their keys in a BST.

2. Step 2: Define inorder successor

   The inorder successor of a node is the next node visited after it in inorder traversal, which is the smallest node with a key greater than the current node's key.

3. Final Answer:

   The node with the smallest key greater than the given node's key -> Option B

4. Quick Check:

   Inorder successor = next larger node [OK]

Hint: Inorder successor is next bigger node in BST [OK]

Common Mistakes:

• Confusing successor with predecessor
• Thinking successor is always right child
• Assuming successor is parent node

2. Which of the following Go function signatures correctly represents a function to find the inorder successor of a BST node?

easy

A. func inorderSuccessor(root *Node, p *Node) *Node

B. func inorderSuccessor(p *Node) *Node

C. func inorderSuccessor(root Node, p Node) Node

D. func inorderSuccessor(root *Node) int

Solution

1. Step 1: Analyze function parameters

   To find the inorder successor, we need the BST root and the target node pointer, both as pointers to Node structs.

2. Step 2: Check return type

   The function should return a pointer to the successor node, so return type is *Node.

3. Final Answer:

   func inorderSuccessor(root *Node, p *Node) *Node -> Option A

4. Quick Check:

   Function needs root and target node pointers, returns node pointer [OK]

Hint: Function needs root and target node pointers [OK]

Common Mistakes:

• Passing nodes by value instead of pointer
• Returning int instead of node pointer
• Missing root parameter

3. Given the following BST and code snippet, what will be printed?

type Node struct {
    Val   int
    Left  *Node
    Right *Node
}

func inorderSuccessor(root, p *Node) *Node {
    var successor *Node
    for root != nil {
        if p.Val < root.Val {
            successor = root
            root = root.Left
        } else {
            root = root.Right
        }
    }
    return successor
}

// BST structure:
//       20
//      /  \
//    10    30
//      \
//      15

p := &Node{Val: 15}
succ := inorderSuccessor(root, p)
if succ != nil {
    fmt.Println(succ.Val)
} else {
    fmt.Println("nil")
}

medium

A. 20

B. 30

C. 15

D. nil

Solution

1. Step 1: Trace inorderSuccessor function for p=15

   Start at root=20: 15 < 20, successor=20, move left to 10. At 10: 15 >= 10, move right to 15. At 15: 15 >= 15, move right (nil). Loop ends.

2. Step 2: Return successor node

   Successor is 20, so succ.Val = 20 is printed.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Next bigger node after 15 is 20 [OK]

Hint: Successor is smallest node bigger than p [OK]

Common Mistakes:

• Returning p itself as successor
• Returning nil when right subtree is empty
• Confusing successor with predecessor

4. The following Go code tries to find the inorder successor of a node in a BST but has a bug. What is the bug?

func inorderSuccessor(root, p *Node) *Node {
    if p.Right != nil {
        curr := p.Right
        for curr.Left != nil {
            curr = curr.Left
        }
        return curr
    }
    var successor *Node
    for root != nil {
        if p.Val < root.Val {
            successor = root
            root = root.Left
        } else {
            root = root.Right
        }
    }
    return successor
}

medium

A. It does not handle the case when p has no right child

B. It incorrectly returns p.Right when p.Right is nil

C. It does not update successor when p.Val equals root.Val

D. It assumes root is always the BST root, which may not be true

Solution

1. Step 1: Understand the logic for successor search

   The code tries to find successor by checking right subtree or walking from root to find successor.

2. Step 2: Identify the bug in root traversal

   The code assumes root is the BST root, but if root is not the actual BST root or p is not reachable from root, the successor found may be incorrect.

3. Final Answer:

   It assumes root is always the BST root, which may not be true -> Option D

4. Quick Check:

   Root must be BST root to find correct successor [OK]

Hint: Ensure root is BST root, not arbitrary node [OK]

Common Mistakes:

• Passing a root that is not the BST root
• Not checking if p.Right is nil before access
• Ignoring case when p has no right subtree

5. You have a BST where each node has a parent pointer in addition to left and right children. How can you find the inorder successor of a node p efficiently using this parent pointer? Choose the correct approach.

hard

A. Always move to the root and perform full inorder traversal to find successor

B. If p has left child, successor is rightmost node in left subtree; else move down to left child

C. If p has right child, successor is leftmost node in right subtree; else move up using parent until node is left child of its parent

D. Successor is always the parent node regardless of children

Solution

1. Step 1: Check if p has right child

   If yes, successor is the leftmost node in p's right subtree.

2. Step 2: If no right child, move up using parent pointer

   Keep moving up until current node is left child of its parent; that parent is the successor.

3. Final Answer:

   If p has right child, successor is leftmost node in right subtree; else move up using parent until node is left child of its parent -> Option C

4. Quick Check:

   Use right subtree or climb parents to find successor [OK]

Hint: Use right subtree or climb parents via left child check [OK]

Common Mistakes:

• Confusing left and right subtree roles
• Always returning parent as successor
• Performing full traversal instead of using parent pointer