BST Inorder Successor in DSA C++ - Execution Trace

Concept Flow - BST Inorder Successor

Start at root

↓

Find node with given key

↓

Does node have right child?

Yes→Go to right child

↓

Find leftmost node in right subtree

↓

Go up to parent while node is right child

↓

Return parent as successor or NULL if none

Find the node, then if it has right child, successor is leftmost in right subtree; else go up to find first ancestor where node is in left subtree.

Execution Sample

DSA C++

Node* inorderSuccessor(Node* root, Node* p) {
  if (p->right) {
    p = p->right;
    while (p->left) p = p->left;
    return p;
  }
  Node* succ = NULL;
  while (root) {
    if (p->val < root->val) {
      succ = root;
      root = root->left;
    } else root = root->right;
  }
  return succ;
}

Finds the inorder successor of node p in BST by checking right subtree or climbing up ancestors.

Execution Table

Step	Operation	Current Node	Pointer Changes	Visual State
1	Start at root to find node 5	10	root=10	10 / \ 5 15
2	Go left since 5 < 10	5	root=5	10 / \ 5 15
3	Node 5 found, check right child	5	p=5	10 / \ 5 15
4	Node 5 has right child 7, go right	7	p=7	10 / \ 5 15 \ 7
5	Find leftmost in right subtree	7	p=7	10 / \ 5 15 \ 7
6	7 has no left child, successor is 7	7	return 7	10 / \ 5 15 \ 7
7	Return successor 7	7	return	Successor: 7
8	Alternate case: Node 15 with no right child	15	p=15	10 / \ 5 15
9	Go up to parent while node is right child	15	root=10	10 / \ 5 15
10	15 is right child of 10, move up	10	root=NULL	10 / \ 5 15
11	No ancestor where node is left child, successor NULL	NULL	return NULL	Successor: NULL

💡 Traversal ends when successor found or no ancestor left where node is left child.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 9	After Step 11
root	10	5	7	7	10	NULL
p	5	5	7	7	15	15
succ	NULL	NULL	NULL	7	NULL	NULL

Key Moments - 3 Insights

Why do we check the right child of the node first?

Because if the node has a right child, the inorder successor is the leftmost node in that right subtree, as shown in steps 3-6 of the execution_table.

What if the node has no right child? How do we find the successor?

We move up the tree to find the first ancestor where the node is in the left subtree, as shown in steps 8-11 where we climb from node 15 up to root 10 and find no successor.

Why do we stop when root becomes NULL in the upward traversal?

Because it means we reached the top of the tree without finding an ancestor where the node is a left child, so the node has no inorder successor (step 11).

Visual Quiz - 3 Questions

Look at the execution_table, what is the successor of node 5 at step 6?

A Node 7

B Node 10

C Node 15

D NULL

💡 Hint

Check the Visual State and Operation columns at step 6 where successor is returned as 7.

At which step does the traversal end with no successor found for node 15?

A Step 7

B Step 9

C Step 11

D Step 4

💡 Hint

Look at the exit_note and rows 10-11 where root becomes NULL and successor is NULL.

If node 5 had no right child, which operation would the algorithm perform next?

A Return node 7 as successor

B Go up to parent to find ancestor where node is left child

C Return NULL immediately

D Go to left child

💡 Hint

Refer to key_moments and execution_table steps 8-11 for the case when right child is absent.

Concept Snapshot

BST Inorder Successor:
- If node has right child, successor = leftmost node in right subtree
- Else, climb ancestors to find first node where current node is in left subtree
- Return NULL if no such ancestor
- Used in BST traversal and deletion
- Time: O(h) where h = tree height

Full Transcript

To find the inorder successor of a node in a BST, first check if the node has a right child. If yes, the successor is the leftmost node in that right subtree. If no right child exists, move up the tree to find the first ancestor where the node is in the left subtree. If none found, successor is NULL. This process is shown step-by-step in the execution table with pointer changes and visual tree states. Key moments clarify why we check right child first and how upward traversal works. The visual quiz tests understanding of these steps.