BST Inorder Predecessor in DSA C++ - Execution Trace

Concept Flow - BST Inorder Predecessor

Start at root node

↓

Search for target node

↓

Found target node?

No→Return None

Yes↓

Does target have left child?

Yes→Go to left child

↓

Go to rightmost node in left subtree

↓

Go up to ancestor where target is in right subtree

↓

Return that ancestor as predecessor

↓

Done

Start from root to find the target node. If it has a left child, predecessor is the rightmost node in left subtree. Otherwise, move up to find the ancestor where target is in right subtree.

Execution Sample

DSA C++

Node* inorderPredecessor(Node* root, Node* target) {
  if (!target) return nullptr;
  if (target->left) {
    Node* curr = target->left;
    while (curr->right) curr = curr->right;
    return curr;
  }
  Node* pred = nullptr;
  Node* curr = root;
  while (curr) {
    if (target->data > curr->data) {
      pred = curr;
      curr = curr->right;
    } else if (target->data < curr->data) {
      curr = curr->left;
    } else break;
  }
  return pred;
}

Finds the inorder predecessor of a target node in a BST by checking left subtree or ancestors.

Execution Table

Step	Operation	Current Node	Pointer Changes	Visual State
1	Start search for target 15	10 (root)	pred=nullptr, curr=10	10 / \ 5 15 / \ 12 20
2	15 > 10, move right	15	pred=10, curr=15	10 / \ 5 15 / \ 12 20
3	Found target 15	15	pred=10, curr=15	10 / \ 5 15 / \ 12 20
4	Target has left child 12	15	curr=12	10 / \ 5 15 / \ 12 20
5	Go to rightmost in left subtree	12	curr=12	10 / \ 5 15 / \ 12 20
6	12 has no right child, stop	12	pred=12	10 / \ 5 15 / \ 12 20
7	Return predecessor 12	12	Return 12	10 / \ 5 15 / \ 12 20
8	Example: target 5 no left child	10	pred=nullptr, curr=10	10 / \ 5 15 / \ 12 20
9	5 < 10, move left	5	pred=nullptr, curr=5	10 / \ 5 15 / \ 12 20
10	Found target 5	5	pred=nullptr, curr=5	10 / \ 5 15 / \ 12 20
11	No left child, move up to find ancestor	5	pred=nullptr, curr=10	10 / \ 5 15 / \ 12 20
12	5 < 10, move left	5	pred=nullptr, curr=5	10 / \ 5 15 / \ 12 20
13	No ancestor where target is in right subtree	null	Return nullptr	10 / \ 5 15 / \ 12 20
14	End	null	Stop	Predecessor is nullptr (no predecessor)

💡 Search ends when target found and predecessor identified or none exists.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	Final
curr	10	15	12	12	12
pred	nullptr	10	10	12	12
target	15	15	15	15	15

Key Moments - 3 Insights

Why do we check the left child of the target node first?

Because if the target has a left subtree, the inorder predecessor is the rightmost node in that left subtree, as shown in steps 4-6 in the execution_table.

What if the target node has no left child? How do we find the predecessor?

We move up the tree from the root, tracking the last node smaller than the target (pred). This is shown in steps 8-13 where no left child exists.

Why do we update 'pred' only when moving right from current node?

Because moving right means current node is less than target, so it could be a predecessor candidate. This is shown in steps 2 and 11.

Visual Quiz - 3 Questions

Look at the execution_table, what is the predecessor when target is 15 at step 7?

A Node with value 12

B Node with value 10

C Node with value 5

D nullptr

💡 Hint

Check step 7 in execution_table where predecessor is returned as 12.

At which step does the search realize the target 5 has no inorder predecessor?

A Step 9

B Step 11

C Step 13

D Step 14

💡 Hint

Look at step 13 where it returns nullptr because no ancestor is found.

If the target node had no left child but was greater than root, how would 'pred' change during search?

A 'pred' would remain nullptr

B 'pred' would update to root or nodes smaller than target

C 'pred' would update to nodes greater than target

D 'pred' would be set to target node

💡 Hint

Refer to variable_tracker and steps where 'pred' updates when moving right.

Concept Snapshot

BST Inorder Predecessor:
- If target has left child, predecessor = rightmost node in left subtree.
- Else, predecessor = ancestor where target is in right subtree.
- Search from root to find target and track predecessor.
- Return nullptr if no predecessor found.
- Useful for BST deletion and traversal.

Full Transcript

To find the inorder predecessor of a node in a BST, first locate the target node starting from the root. If the target has a left child, the predecessor is the rightmost node in that left subtree. If no left child exists, move up the tree from the root, tracking the last node smaller than the target, which becomes the predecessor. If no such ancestor exists, the predecessor is null. This method helps in BST operations like deletion and inorder traversal.

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 visited just before it in 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 B

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Predecessor is just before in sorted order [OK]

Common Mistakes:

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

2. Which of the following C++ function signatures correctly represents a function to find the inorder predecessor of a BST node given its pointer Node* root and Node* target?

easy

A. Node* findInorderPredecessor(Node* root, Node* target);

B. void findInorderPredecessor(Node* root, int target);

C. int findInorderPredecessor(Node* root, Node* target);

D. Node findInorderPredecessor(Node root, Node target);

Solution

1. Step 1: Analyze parameter types

   The function needs the root pointer and the target node pointer to find the predecessor.

2. Step 2: Analyze return type

   The predecessor is a node pointer, so return type should be Node*.

3. Final Answer:

   Node* findInorderPredecessor(Node* root, Node* target); -> Option A

4. Quick Check:

   Correct parameters and return type [OK]

Hint: Return pointer to predecessor node [OK]

Common Mistakes:

• Using int return type instead of Node*
• Passing target as int instead of Node*
• Passing nodes by value instead of pointer

3. Consider the following BST and code snippet to find the inorder predecessor of node with value 15:

struct Node {
  int val;
  Node* left;
  Node* right;
};

Node* findMax(Node* root) {
  while (root->right) root = root->right;
  return root;
}

Node* inorderPredecessor(Node* root, Node* target) {
  if (target->left) return findMax(target->left);
  Node* predecessor = nullptr;
  while (root) {
    if (target->val > root->val) {
      predecessor = root;
      root = root->right;
    } else if (target->val < root->val) {
      root = root->left;
    } else break;
  }
  return predecessor;
}

// BST structure:
//        20
//       /  \
//     10    30
//       \     \
//       15    40

Node* root = ...; // root of above BST
Node* target = root->left->right; // node with value 15
Node* pred = inorderPredecessor(root, target);
std::cout << pred->val << std::endl;

What is the output?

medium

A. 10

B. 15

C. 20

D. 30

Solution

1. Step 1: Check if target has left subtree

   Node 15 has no left child, so we skip findMax on left subtree.

2. Step 2: Traverse from root to find predecessor

   Start at root (20). Since 15 < 20, go left to 10. Since 15 > 10, set predecessor = 10 and go right to 15. Found target, stop.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Largest smaller value than 15 is 10 [OK]

Hint: If no left child, track last smaller ancestor [OK]

Common Mistakes:

• Returning target node itself
• Returning root node 20 incorrectly
• Confusing predecessor with successor

4. The following code attempts to find the inorder predecessor of a node in a BST but has a bug. Identify the bug.

Node* inorderPredecessor(Node* root, Node* target) {
  if (target->left) {
    Node* temp = target->left;
    while (temp->left) temp = temp->left;
    return temp;
  }
  Node* predecessor = nullptr;
  while (root) {
    if (target->val > root->val) {
      predecessor = root;
      root = root->right;
    } else if (target->val < root->val) {
      root = root->left;
    } else break;
  }
  return predecessor;
}

medium

A. The predecessor variable is not initialized

B. The function does not handle the case when target has no left subtree

C. The function returns nullptr always

D. The loop to find max in left subtree incorrectly goes left instead of right

Solution

1. Step 1: Check left subtree max search

   To find inorder predecessor in left subtree, we must find the maximum node, which is the rightmost node, but code goes left repeatedly.

2. Step 2: Confirm other parts are correct

   The rest of the code correctly handles cases without left subtree and initializes predecessor.

3. Final Answer:

   The loop to find max in left subtree incorrectly goes left instead of right -> Option D

4. Quick Check:

   Max in left subtree requires right traversal [OK]

Hint: Max in left subtree means go right, not left [OK]

Common Mistakes:

• Going left instead of right to find max
• Not checking if target->left is null
• Confusing predecessor with successor

5. Given a BST with nodes inserted in this order: 25, 15, 50, 10, 22, 35, 70, 4, 12, 18, 24, 31, 44, 66, 90. Write a function to find the inorder predecessor of a node with value 31. What will be the predecessor's value? Hint: Use the BST property and consider both cases when the node has a left subtree or not.

hard

A. 25

B. 35

C. 24

D. 22

Solution

1. Step 1: Locate node 31

   Node 31 is the left child of 35 and has no left subtree.

2. Step 2: Traverse from root to find predecessor

   Start at root 25 (31 > 25, predecessor=25, go right to 50). 31 < 50, go left to 35. 31 < 35, go left to 31. Found target, predecessor=25.

3. Step 3: Verify with inorder traversal

   Inorder traversal sequence around 31 is ... 24, 25, 31, 35 ... The predecessor of 31 is the largest smaller value, which is 24 (left subtree of 25).

4. Final Answer:

   24 -> Option C

5. Quick Check:

   Inorder predecessor of 31 is 24 [OK]

Hint: If no left child, predecessor is last smaller ancestor or max in left subtree [OK]

Common Mistakes:

• Choosing 25 instead of 24
• Choosing successor 35 instead of predecessor
• Ignoring left subtree check