BST Inorder Predecessor in DSA Go - Execution Trace

Concept Flow - BST Inorder Predecessor

Start at root

↓

Search for node with given key

↓

Node found?

No→Return nil (no predecessor)

Yes↓

Does node have left child?

Yes→Go to left child

↓

Go to rightmost node in left subtree

↓

This is inorder predecessor

No↓

Go up to ancestor where node is in right subtree

↓

That ancestor is inorder predecessor

↓

Return predecessor node

Find the node with the given key, then find its inorder predecessor by either going to the rightmost node in its left subtree or moving up to the nearest ancestor where the node is in the right subtree.

Execution Sample

DSA Go

func inorderPredecessor(root, key *Node) *Node {
  node := search(root, key.val)
  if node == nil {
    return nil
  }
  if node.left != nil {
    return rightmost(node.left)
  }
  return ancestorPredecessor(root, node)
}

This code finds the inorder predecessor of a node in a BST by searching the node, then checking its left subtree or ancestors.

Execution Table

Step	Operation	Node Visited	Pointer Changes	Visual State
1	Start at root	50	head=root(50)	50 / \ 30 70 / \ / \ 20 40 60 80
2	Search for node with key=40	50	current=50	Same as above
3	Move left (40 < 50)	30	current=30	Same as above
4	Move right (40 > 30)	40	current=40	Same as above
5	Node found	40	node=40	Same as above
6	Check if node.left exists	40	node.left=nil	Same as above
7	No left child, find ancestor predecessor	40	start=root(50)	Same as above
8	Move down from root to find ancestor	50	predecessor=nil	Same as above
9	40 < 50, move left	30	predecessor=nil	Same as above
10	40 > 30, set predecessor=30, move right	40	predecessor=30	Same as above
11	Reached node, stop	40	predecessor=30	Same as above
12	Return predecessor node	30	return 30	Same as above

💡 Node 40 has no left child; predecessor is ancestor 30 found by moving up.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 10	Final
current	root(50)	50	30	40	40	40	40	40
node	nil	nil	nil	nil	40	40	40	40
predecessor	nil	nil	nil	nil	nil	nil	30	30

Key Moments - 3 Insights

Why do we check if the node has a left child first?

Because if the node has a left child, its inorder predecessor is the rightmost node in that left subtree (see step 6 and 7). If no left child, we look for ancestor predecessor.

Why do we move up to ancestors when there is no left child?

Because the inorder predecessor is the nearest ancestor where the node lies in the right subtree (see steps 7 to 11). This ancestor is the predecessor.

Why do we stop searching ancestor when we reach the node itself?

Because once we reach the node during ancestor traversal, we have found the predecessor stored in 'predecessor' variable (step 11).

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'predecessor' after step 10?

A 30

B 40

C 50

D nil

💡 Hint

Check the 'predecessor' column in variable_tracker after step 10.

At which step do we confirm that the node has no left child?

A Step 5

B Step 6

C Step 7

D Step 10

💡 Hint

Look at the 'Operation' column in execution_table where left child check happens.

If the node had a left child, which step would we go to next?

A Step 7

B Step 6

C Step 4

D Step 12

💡 Hint

Refer to the decision at step 6 in execution_table about left child presence.

Concept Snapshot

BST Inorder Predecessor:
- Find node with given key.
- If node has left child, predecessor is rightmost node in left subtree.
- Else, predecessor is nearest ancestor where node is in right subtree.
- Return predecessor node or nil if none.
- Used in BST deletion and traversal.

Full Transcript

To find the inorder predecessor in a BST, first search for the node with the given key starting from the root. If the node is found and has a left child, the predecessor is the rightmost node in its left subtree. If there is no left child, move up the tree to find the nearest ancestor where the node lies in the right subtree; that ancestor is the predecessor. If no such ancestor exists, the node has no inorder predecessor. This process is shown step-by-step in the execution table and variable tracker.

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