BST Inorder Predecessor in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum value node in the left subtree.

DSA Rust

fn max_node(mut node: Option<&Box<TreeNode>>) -> Option<&Box<TreeNode>> {
    while let Some(ref current) = node {
        if current.right.is_none() {
            return node;
        }
        node = current.[1].as_ref();
    }
    None
}

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

Aleft

Bvalue

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

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

Trying to access 'parent' which is not stored in this node structure.

To find the maximum node in a subtree, we keep moving to the right child until there is no right child.

2fill in blank

medium

Complete the code to move to the left child of the root to find the inorder predecessor.

DSA Rust

fn inorder_predecessor(root: Option<&Box<TreeNode>>, key: i32) -> Option<&Box<TreeNode>> {
    let mut current = root;
    while let Some(node) = current {
        if node.value == key {
            if node.left.is_some() {
                return max_node(node.[1].as_ref());
            } else {
                break;
            }
        } else if key < node.value {
            current = node.left.as_ref();
        } else {
            current = node.right.as_ref();
        }
    }
    None
}

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' will lead to wrong subtree.

Trying to access 'parent' which is not stored.

The inorder predecessor is the maximum node in the left subtree, so we move to the left child first.

3fill in blank

hard

Fix the error in updating the predecessor when the node has no left child.

DSA Rust

fn inorder_predecessor(root: Option<&Box<TreeNode>>, key: i32) -> Option<&Box<TreeNode>> {
    let mut predecessor = None;
    let mut current = root;
    while let Some(node) = current {
        if key > node.value {
            predecessor = current;
            current = node.[1].as_ref();
        } else {
            current = node.left.as_ref();
        }
    }
    predecessor
}

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

Aleft

Bparent

Cright

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' will cause infinite loop or wrong traversal.

Trying to access 'parent' which is not stored.

When key is greater than node value, move to the right child to find a closer predecessor candidate.

4fill in blank

hard

Fill both blanks to complete the function that finds the inorder predecessor when the node has no left subtree.

DSA Rust

fn inorder_predecessor(root: Option<&Box<TreeNode>>, key: i32) -> Option<&Box<TreeNode>> {
    let mut predecessor = None;
    let mut current = root;
    while let Some(node) = current {
        if key > node.value {
            predecessor = current;
            current = node.[1].as_ref();
        } else {
            current = node.[2].as_ref();
        }
    }
    predecessor
}

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

Aright

Bleft

Cparent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right pointers causes wrong traversal.

Trying to access 'parent' which is not stored.

When key is greater, move right to find a closer predecessor; else move left to continue search.

5fill in blank

hard

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

DSA Rust

fn inorder_predecessor(root: Option<&Box<TreeNode>>, key: i32) -> Option<&Box<TreeNode>> {
    let mut current = root;
    while let Some(node) = current {
        if node.value == key {
            if node.left.is_some() {
                return max_node(node.[1].as_ref());
            } else {
                break;
            }
        } else if key < node.value {
            current = node.[2].as_ref();
        } else {
            current = node.[3].as_ref();
        }
    }
    let mut predecessor = None;
    current = root;
    while let Some(node) = current {
        if key > node.value {
            predecessor = current;
            current = node.right.as_ref();
        } else {
            current = node.left.as_ref();
        }
    }
    predecessor
}

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

Aleft

Bright

Cparent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' which is not stored.

Swapping left and right pointers.

The predecessor is max node in left subtree if left exists; else search path updating predecessor when moving right.

Practice

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

easy

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

B. The node with the smallest value larger 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 BST property

   In a BST, left subtree nodes are smaller and right subtree nodes are larger than the current node.

2. Step 2: Define inorder predecessor

   The inorder predecessor is the node visited just before the current node in an inorder traversal, which means it has the largest value smaller than the current node.

3. Final Answer:

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

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Inorder predecessor is largest smaller node [OK]

Common Mistakes:

• Confusing predecessor with successor
• Choosing child nodes directly without checking values
• Thinking predecessor is always left child

2. Which Rust code snippet correctly finds the inorder predecessor of a node with a left subtree in a BST?

easy

A. let pred = node.right.clone();

B. let mut pred = node.right.as_ref(); while let Some(left) = pred.unwrap().left.as_ref() { pred = Some(left); }

C. let mut pred = node.left.as_ref(); while let Some(right) = pred.unwrap().right.as_ref() { pred = Some(right); }

D. let pred = node.left.clone();

Solution

1. Step 1: Identify left subtree traversal

   The inorder predecessor is the rightmost node in the left subtree, so start from node.left.

2. Step 2: Traverse to rightmost node

   Keep moving to the right child until no more right child exists to find the largest node in left subtree.

3. Final Answer:

   let mut pred = node.left.as_ref(); while let Some(right) = pred.unwrap().right.as_ref() { pred = Some(right); } -> Option C

4. Quick Check:

   Traverse left subtree rightmost node = let mut pred = node.left.as_ref(); while let Some(right) = pred.unwrap().right.as_ref() { pred = Some(right); } [OK]

Hint: Go left once, then right as far as possible [OK]

Common Mistakes:

• Starting from right subtree instead of left
• Not looping to the rightmost node
• Using clone instead of reference traversal

3. Given the BST below and node value 15, what is the inorder predecessor?

       20
      /  \
    10    30
      \     \
      15    40

medium

A. 30

B. 15

C. 20

D. 10

Solution

1. Step 1: Locate node 15 and its left subtree

   Node 15 has no left child, so we look for the ancestor smaller than 15.

2. Step 2: Find largest ancestor smaller than 15

   10 is the parent of 15 and smaller than 15, so 10 is the inorder predecessor.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Largest smaller ancestor = 10 [OK]

Hint: If no left child, go up to smaller ancestor [OK]

Common Mistakes:

• Choosing node itself as predecessor
• Choosing larger ancestor incorrectly
• Ignoring no left subtree case

4. Identify the bug in this Rust function to find the inorder predecessor of a BST node:

fn inorder_predecessor(node: &Option) -> Option<i32> {
    let mut current = node.as_ref()?.left.as_ref();
    while current?.right.is_some() {
        current = current?.right.as_ref();
    }
    current.map(|n| n.val)
}

medium

A. Using '?' operator on Option without handling None case

B. Starting from node.right instead of node.left

C. Returning node.val instead of predecessor value

D. Not checking if node has left child before traversal

Solution

1. Step 1: Analyze use of '?' operator in loop condition

   The '?' operator unwraps Option or returns None; using it inside while condition causes panic if None.

2. Step 2: Identify safe way to traverse Option references

   Must use pattern matching or if let to safely unwrap Options instead of '?' in loop condition.

3. Final Answer:

   Using '?' operator on Option without handling None case -> Option A

4. Quick Check:

   Unsafe '?' in loop condition causes panic [OK]

Hint: Avoid '?' in loop conditions on Option [OK]

Common Mistakes:

• Using '?' inside loops causing runtime panic
• Ignoring None cases in Option traversal
• Confusing left and right subtree

5. You want to find the inorder predecessor of a node in a BST that has no left child. Which approach correctly finds it?

hard

A. Return the rightmost node in the right subtree

B. Traverse up the tree to find the nearest ancestor whose right child is also an ancestor of the node

C. Return the leftmost node in the right subtree

D. Return the root node always

Solution

1. Step 1: Understand no left child case

   If node has no left child, predecessor is not in left subtree but among ancestors.

2. Step 2: Find nearest ancestor where node is in right subtree

   Traverse up until you find an ancestor whose right child is also an ancestor of the node; that ancestor is the predecessor.

3. Final Answer:

   Traverse up the tree to find the nearest ancestor whose right child is also an ancestor of the node -> Option B

4. Quick Check:

   No left child -> find right-side ancestor [OK]

Hint: Go up to ancestor where node is in right subtree [OK]

Common Mistakes:

• Returning nodes from right subtree incorrectly
• Always returning root node
• Ignoring ancestor traversal