Tree Traversal Postorder Left Right Root in DSA C++ - Execution Trace

Concept Flow - Tree Traversal Postorder Left Right Root

Start at root node

↓

Traverse left subtree

↓

Traverse right subtree

↓

Visit current node (root)

↓

Back to parent node or end

The traversal visits left subtree first, then right subtree, and finally the root node.

Execution Sample

DSA C++

void postorder(Node* root) {
  if (!root) return;
  postorder(root->left);
  postorder(root->right);
  cout << root->val << " ";
}

This code prints nodes of a binary tree in postorder: left, right, then root.

Execution Table

Step	Operation	Node Visited	Pointer Changes	Tree State (Visual)
1	Start at root	A	current = A	A / \ B C
2	Traverse left subtree	B	current = B	A / \ B C
3	Traverse left subtree of B	D	current = D	A / \ B C / D
4	Left of D is null	null	return	A / \ B C / D
5	Right of D is null	null	return	A / \ B C / D
6	Visit node D	D	print D	A / \ B C / D
7	Traverse right subtree of B	E	current = E	A / \ B C / \ D E
8	Left of E is null	null	return	A / \ B C / \ D E
9	Right of E is null	null	return	A / \ B C / \ D E
10	Visit node E	E	print E	A / \ B C / \ D E
11	Visit node B	B	print B	A / \ B C / \ D E
12	Traverse right subtree	C	current = C	A / \ B C
13	Left of C is null	null	return	A / \ B C
14	Right of C is null	null	return	A / \ B C
15	Visit node C	C	print C	A / \ B C
16	Visit root node	A	print A	A / \ B C
17	Traversal complete	null	return	Traversal done

💡 All nodes visited in postorder: left, right, root; traversal ends.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 6	After Step 10	After Step 11	After Step 15	After Step 16	Final
current	null	A	B	D	D	E	B	C	A	null
printed_nodes	[]	[]	[]	[]	[D]	[D, E]	[D, E, B]	[D, E, B, C]	[D, E, B, C, A]	[D, E, B, C, A]

Key Moments - 3 Insights

Why do we print the node after traversing both left and right subtrees?

Because postorder means visiting left, then right, then the root node. The execution_table rows 4-6 and 8-10 show traversal of children before printing.

What happens when a node has no left or right child?

The function returns immediately without printing, as shown in steps 4, 5, 8, 9, 13, and 14 where null nodes cause return.

Why does the traversal start at the root but print it last?

Traversal visits children first recursively before printing the root, as seen in steps 1 (start at A) and 16 (print A last).

Visual Quiz - 3 Questions

Look at the execution_table, what nodes have been printed after step 10?

A [D, E, B]

B [D]

C [D, E]

D []

💡 Hint

Check the 'printed_nodes' variable in variable_tracker after step 10.

At which step does the traversal visit the root node 'A'?

A Step 16

B Step 6

C Step 11

D Step 15

💡 Hint

Look at the 'Operation' and 'Node Visited' columns in execution_table for when 'A' is printed.

If node 'B' had no right child, how would the printed nodes after step 10 change?

A [D]

B [D, B]

C [D, E]

D [D, E, B]

💡 Hint

Without right child 'E', node 'B' would be printed after left subtree only, see steps 7-11.

Concept Snapshot

Postorder traversal visits nodes in Left → Right → Root order.
Use recursion: traverse left subtree, then right subtree, then print root.
Base case: if node is null, return immediately.
Useful for deleting trees or postfix expressions.
Output order for example tree: D E B C A

Full Transcript

Postorder traversal means visiting the left child first, then the right child, and finally the root node. The code uses recursion to do this. It starts at the root node, goes down to the leftmost node, prints it, then moves to the right sibling, prints it, and finally prints the parent node. This repeats up the tree until all nodes are printed. If a node is null, the function returns immediately without printing. The printed order for the example tree is D, E, B, C, A. This traversal is useful for tasks like deleting a tree or evaluating postfix expressions.