BST Inorder Predecessor in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the left child of a BST node.

DSA Go

func getLeftChild(node *Node) *Node {
    return node.[1]
}

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

ALeft

BRight

CParent

DValue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Right' instead of 'Left' to get the left child.

Trying to access 'Value' instead of child nodes.

The left child of a BST node is accessed using the 'Left' field.

2fill in blank

medium

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

DSA Go

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

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

AParent

BRight

CLeft

DValue

Attempts:

3 left

💡 Hint

Common Mistakes

Going left instead of right to find maximum.

Checking 'Value' instead of child pointers.

The maximum node in a BST subtree is found by going right until no more right child exists.

3fill in blank

hard

Fix the error in the code to find the inorder predecessor when left subtree exists.

DSA Go

func inorderPredecessor(node *Node) *Node {
    if node.Left != nil {
        pred := node.Left
        for pred.[1] != nil {
            pred = pred.[1]
        }
        return pred
    }
    return nil
}

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

AValue

BParent

CLeft

DRight

Attempts:

3 left

💡 Hint

Common Mistakes

Going left instead of right inside left subtree.

Returning the left child directly without traversal.

To find the inorder predecessor in the left subtree, go to the rightmost node of the left child.

4fill in blank

hard

Fill both blanks to find the inorder predecessor when no left subtree exists.

DSA Go

func inorderPredecessor(node *Node) *Node {
    if node.Left == nil {
        pred := node.[1]
        for pred != nil && node == pred.[2] {
            node = pred
            pred = pred.[1]
        }
        return pred
    }
    return nil
}

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

AParent

BLeft

CRight

DValue

Attempts:

3 left

💡 Hint

Common Mistakes

Using Right instead of Left to check child relation.

Not moving up the tree using Parent pointer.

When no left subtree, move up using Parent pointer until node is not a left child.

5fill in blank

hard

Fill all three blanks to complete the full inorder predecessor function.

DSA Go

func inorderPredecessor(node *Node) *Node {
    if node.Left != nil {
        pred := node.Left
        for pred.[1] != nil {
            pred = pred.[1]
        }
        return pred
    }
    pred := node.[2]
    for pred != nil && node == pred.[3] {
        node = pred
        pred = pred.[2]
    }
    return pred
}

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

ARight

BParent

CLeft

DValue

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing Left and Right pointers incorrectly.

Not handling both cases of predecessor properly.

If left subtree exists, go rightmost in left subtree. Else, move up parents until node is not left child.

Practice

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

easy

A. The node with the smallest value larger than the given node

B. The node with the largest value smaller than the given node

C. The left child of the given node

D. The right child of the given node

Solution

1. Step 1: Understand inorder traversal in BST

   Inorder traversal visits nodes in ascending order of their values.

2. Step 2: Define inorder predecessor

   The inorder predecessor of a node is the node that comes just before it in this ascending order, meaning it has the largest value smaller than the node.

3. Final Answer:

   The node with the largest value smaller than the given node -> Option B

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Predecessor is largest smaller value in BST [OK]

Common Mistakes:

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

2. Which of the following Go function signatures correctly represents a function to find the inorder predecessor in a BST node struct named Node?

easy

A. func inorderPredecessor(root *Node, key int) *Node

B. func inorderPredecessor(key int) Node

C. func inorderPredecessor(root Node, key int) Node

D. func inorderPredecessor(root *Node) int

Solution

1. Step 1: Analyze function parameters and return type

   The function needs the BST root pointer and the key to find predecessor for, and returns a pointer to the predecessor node.

2. Step 2: Match with options

   func inorderPredecessor(root *Node, key int) *Node matches this: it takes *Node root, int key, and returns *Node.

3. Final Answer:

   func inorderPredecessor(root *Node, key int) *Node -> Option A

4. Quick Check:

   Pointer root and key input, pointer node output [OK]

Hint: Use pointer receiver and return pointer for BST nodes [OK]

Common Mistakes:

• Passing root by value instead of pointer
• Returning node value instead of node pointer
• Missing key parameter to identify target node

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

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

func inorderPredecessor(root *Node, key int) *Node {
    var predecessor *Node
    curr := root
    for curr != nil {
        if key <= curr.Val {
            curr = curr.Left
        } else {
            predecessor = curr
            curr = curr.Right
        }
    }
    return predecessor
}

func main() {
    root := &Node{Val: 20}
    root.Left = &Node{Val: 10}
    root.Right = &Node{Val: 30}
    root.Left.Left = &Node{Val: 5}
    root.Left.Right = &Node{Val: 15}

    pred := inorderPredecessor(root, 15)
    if pred != nil {
        fmt.Println(pred.Val)
    } else {
        fmt.Println("No predecessor")
    }
}

medium

A. No predecessor

B. 5

C. 15

D. 10

Solution

1. Step 1: Trace the inorderPredecessor function for key=15

   Start at root (20). Since 15 <= 20, move left to 10. Since 15 > 10, set predecessor=10, move right to 15. Since 15 <= 15, move left (nil). Loop ends.

2. Step 2: Return predecessor node value

   Predecessor is node with value 10, so print 10.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Largest smaller than 15 is 10 [OK]

Hint: Move left if key ≤ node, else update predecessor and move right [OK]

Common Mistakes:

• Returning the key itself as predecessor
• Confusing predecessor with successor
• Not updating predecessor correctly when moving right

4. The following Go function is intended to find the inorder predecessor of a node with value key in a BST. What is the bug?

func inorderPredecessor(root *Node, key int) *Node {
    var predecessor *Node
    curr := root
    for curr != nil {
        if key < curr.Val {
            curr = curr.Left
        } else {
            predecessor = curr
            curr = curr.Right
        }
    }
    return predecessor
}

medium

A. The function should return curr instead of predecessor

B. The predecessor variable should be reset inside the loop

C. The condition should be key <= curr.Val to handle equal keys correctly

D. The loop should use recursion instead of iteration

Solution

1. Step 1: Analyze the condition in the loop

   The current condition is key < curr.Val, which skips updating predecessor when key equals curr.Val.

2. Step 2: Correct the condition to include equality

   Changing to key <= curr.Val ensures that when key equals curr.Val, we move left without updating predecessor, which is correct for inorder predecessor logic.

3. Final Answer:

   The condition should be key <= curr.Val to handle equal keys correctly -> Option C

4. Quick Check:

   Use <= to handle equal keys properly [OK]

Hint: Use <= in condition to handle equal keys correctly [OK]

Common Mistakes:

• Using < instead of <= causes wrong predecessor for equal keys
• Resetting predecessor inside loop loses previous valid value
• Returning curr instead of predecessor returns wrong node

5. You have a BST where each node has a pointer to its parent. How would you find the inorder predecessor of a given node n efficiently without starting from the root? Choose the best approach.

hard

A. If n has a left child, find the rightmost node in its left subtree; else, move up using parent pointers until you find a node which is the right child of its parent

B. Always start from the root and do a full inorder traversal to find predecessor

C. If n has a right child, find the leftmost node in its right subtree; else, move up using parent pointers until you find a node which is the left child of its parent

D. Return the parent node directly as the predecessor

Solution

1. Step 1: Check if node has left subtree

   If n has a left child, the predecessor is the rightmost node in that left subtree.

2. Step 2: If no left child, move up using parent pointers

   Move up until you find a node that is the right child of its parent; that parent is the predecessor.

3. Final Answer:

   If n has left child, find rightmost in left subtree; else move up until node is right child -> Option A

4. Quick Check:

   Use left subtree or climb parents to find predecessor [OK]

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

Common Mistakes:

• Starting from root wastes time
• Confusing predecessor with successor logic
• Returning parent directly without checks