BST Inorder Successor in DSA Go - Build from Scratch

BST Inorder Successor

📖 Scenario: You are working with a Binary Search Tree (BST) that stores numbers. You want to find the next bigger number after a given number in the tree. This is called the inorder successor.Think of the BST as a sorted list of numbers arranged in a tree shape. Finding the inorder successor means finding the number that comes right after a given number when you look at the numbers in order.

🎯 Goal: Build a program in Go that creates a BST, sets a target node value, finds the inorder successor of that node, and prints the result.

📋 What You'll Learn

Create a BST with the exact nodes: 20, 9, 25, 5, 12, 11, 14

Set a variable called target to the node with value 9

Write a function inorderSuccessor that finds the inorder successor of the target node in the BST

Print the value of the inorder successor node or nil if none exists

💡 Why This Matters

🌍 Real World

Finding the inorder successor in a BST is useful in database indexing, where you want to find the next record in sorted order efficiently.

💼 Career

Understanding BST operations like inorder successor is important for software engineers working with search algorithms, data indexing, and optimization.

1

Create the BST nodes

Create the BST nodes with values 20, 9, 25, 5, 12, 11, and 14 using the TreeNode struct. Connect them to form the BST exactly as follows: 20 is root, 9 is left child of 20, 25 is right child of 20, 5 is left child of 9, 12 is right child of 9, 11 is left child of 12, and 14 is right child of 12.

DSA Go

package main

import "fmt"

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

func main() {
    // Create BST nodes and connect them
    // Your code here
}

Hint

Use &TreeNode{Val: value} to create nodes and assign them to the correct Left or Right fields.

2

Set the target node

Create a variable called target and set it to point to the node with value 9 in the BST.

DSA Go

package main

import "fmt"

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

func main() {
    root := &TreeNode{Val: 20}
    root.Left = &TreeNode{Val: 9}
    root.Right = &TreeNode{Val: 25}
    root.Left.Left = &TreeNode{Val: 5}
    root.Left.Right = &TreeNode{Val: 12}
    root.Left.Right.Left = &TreeNode{Val: 11}
    root.Left.Right.Right = &TreeNode{Val: 14}

    // Set target node to root.Left
    // Your code here
}

Hint

The node with value 9 is the left child of the root node.

3

Write the inorderSuccessor function

Write a function called inorderSuccessor that takes root and target nodes as input and returns the inorder successor node of target in the BST. Use the BST property to find the successor efficiently.

DSA Go

package main

import "fmt"

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

func inorderSuccessor(root *TreeNode, target *TreeNode) *TreeNode {
    // Your code here
}

func main() {
    root := &TreeNode{Val: 20}
    root.Left = &TreeNode{Val: 9}
    root.Right = &TreeNode{Val: 25}
    root.Left.Left = &TreeNode{Val: 5}
    root.Left.Right = &TreeNode{Val: 12}
    root.Left.Right.Left = &TreeNode{Val: 11}
    root.Left.Right.Right = &TreeNode{Val: 14}

    target := root.Left
}

Hint

Use a loop starting from root. If target.Val is less than current.Val, update successor and go left. Otherwise, go right. Return successor at the end.

4

Print the inorder successor value

Call the inorderSuccessor function with root and target. Print the value of the returned node if it is not nil. Otherwise, print nil.

DSA Go

package main

import "fmt"

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

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

func main() {
    root := &TreeNode{Val: 20}
    root.Left = &TreeNode{Val: 9}
    root.Right = &TreeNode{Val: 25}
    root.Left.Left = &TreeNode{Val: 5}
    root.Left.Right = &TreeNode{Val: 12}
    root.Left.Right.Left = &TreeNode{Val: 11}
    root.Left.Right.Right = &TreeNode{Val: 14}

    target := root.Left

    // Print the inorder successor value or nil
    // Your code here
}

Hint

Call inorderSuccessor(root, target). If the result is not nil, print its Val. Otherwise, print nil.

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