Tree Traversal Postorder Left Right Root in DSA Java - Time & Space Complexity

Time Complexity: Tree Traversal Postorder Left Right Root

O(n)

Understanding Time Complexity

We want to understand how long it takes to visit all nodes in a tree using postorder traversal.

How does the time grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void postorderTraversal(TreeNode node) {
    if (node == null) return;
    postorderTraversal(node.left);
    postorderTraversal(node.right);
    System.out.print(node.val + " ");
}
    

This code visits each node in the tree after visiting its left and right children.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

Each node is visited exactly once, so the work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The time grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows in direct proportion to the number of nodes.

Common Mistake

[X] Wrong: "Postorder traversal takes more than linear time because it visits nodes multiple times."

[OK] Correct: Each node is visited only once, even though recursion happens multiple times, so total work is linear.

Interview Connect

Understanding this traversal's time complexity helps you explain how tree algorithms scale, a key skill in many coding challenges.

Self-Check

"What if we changed the traversal to visit nodes in preorder (root left right)? How would the time complexity change?"

Practice

1. What is the correct order of nodes visited in a postorder tree traversal?

easy

A. Root node, Left subtree, Right subtree

B. Left subtree, Right subtree, Root node

C. Root node, Right subtree, Left subtree

D. Right subtree, Left subtree, Root node

Solution

1. Step 1: Understand postorder traversal definition

   Postorder traversal visits nodes in the order: left subtree first, then right subtree, and finally the root node.

2. Step 2: Match the order with options

   Left subtree, Right subtree, Root node matches the order Left, Right, Root exactly.

3. Final Answer:

   Left subtree, Right subtree, Root node -> Option B

4. Quick Check:

   Postorder = Left, Right, Root [OK]

Hint: Remember: postorder ends with root after children [OK]

Common Mistakes:

• Confusing postorder with preorder or inorder
• Visiting root before children
• Swapping left and right subtree order

2. Which of the following Java method signatures correctly represents a recursive postorder traversal of a binary tree node named root?

easy

A. void postorder(Node root) { postorder(root.left); postorder(root.right); visit(root); }

B. void postorder(Node root) { visit(root); postorder(root.left); postorder(root.right); }

C. void postorder(Node root) { postorder(root.right); postorder(root.left); visit(root); }

D. void postorder(Node root) { visit(root.left); visit(root.right); visit(root); }

Solution

1. Step 1: Recall postorder traversal steps

   Postorder visits left subtree, then right subtree, then root node (visit).

2. Step 2: Check each method's order

   void postorder(Node root) { postorder(root.left); postorder(root.right); visit(root); } calls postorder on left, then right, then visits root, matching postorder correctly.

3. Final Answer:

   void postorder(Node root) { postorder(root.left); postorder(root.right); visit(root); } -> Option A

4. Quick Check:

   Left, Right, Root method order = void postorder(Node root) { postorder(root.left); postorder(root.right); visit(root); } [OK]

Hint: Postorder calls left, right, then visit root [OK]

Common Mistakes:

• Visiting root before children
• Swapping left and right recursive calls
• Using visit instead of recursive call for children

3. Given the binary tree below:

    1
   / \
  2   3
 / \
4   5

What is the output of the postorder traversal?

medium

A. [1, 2, 3, 4, 5]

B. [2, 4, 5, 3, 1]

C. [4, 5, 2, 3, 1]

D. [4, 2, 5, 3, 1]

Solution

1. Step 1: Traverse left subtree of root (1)

   Left subtree rooted at 2: postorder visits 4 (left), 5 (right), then 2 (root) -> [4, 5, 2]

2. Step 2: Traverse right subtree and root

   Right subtree is 3 (leaf), so visit 3, then visit root 1 -> [3, 1]

3. Step 3: Combine all parts

   Full postorder: left subtree [4,5,2], right subtree [3], root [1] -> [4, 5, 2, 3, 1]

4. Final Answer:

   [4, 5, 2, 3, 1] -> Option C

5. Quick Check:

   Postorder = Left, Right, Root = [4,5,2,3,1] [OK]

Hint: Visit left subtree fully, then right, then root last [OK]

Common Mistakes:

• Visiting root before children
• Mixing up left and right subtree order
• Outputting preorder or inorder instead

4. Consider this Java method for postorder traversal:

void postorder(Node root) {
  if (root == null) return;
  postorder(root.right);
  postorder(root.left);
  visit(root);
}

What is the main problem with this code?

medium

A. It visits right subtree before left subtree, reversing order

B. It causes infinite recursion

C. It visits the root before children

D. It does not check for null root

Solution

1. Step 1: Analyze traversal order in code

   The code calls postorder on right subtree first, then left subtree, then visits root.

2. Step 2: Compare with correct postorder order

   Correct postorder is left, right, root. This code reverses left and right subtree order.

3. Final Answer:

   It visits right subtree before left subtree, reversing order -> Option A

4. Quick Check:

   Left before right needed, code does right before left [OK]

Hint: Left subtree must be visited before right in postorder [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Ignoring null check (though present here)
• Visiting root before children

5. You want to collect all node values of a binary tree in postorder traversal into a list. Which of the following Java code snippets correctly implements this?

hard

A. void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); result.add(root.val); postorder(root.right, result); }

B. void postorder(Node root, List<Integer> result) { if(root==null) return; result.add(root.val); postorder(root.left, result); postorder(root.right, result); }

C. void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.right, result); postorder(root.left, result); result.add(root.val); }

D. void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); postorder(root.right, result); result.add(root.val); }

Solution

1. Step 1: Understand postorder traversal with list collection

   Postorder visits left subtree, then right subtree, then adds root value to list.

2. Step 2: Check each code snippet's order

   void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); postorder(root.right, result); result.add(root.val); } calls postorder on left, then right, then adds root value last, matching postorder correctly.

3. Final Answer:

   void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); postorder(root.right, result); result.add(root.val); } -> Option D

4. Quick Check:

   Left, Right, Root with result.add last = void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); postorder(root.right, result); result.add(root.val); } [OK]

Hint: Add root value after left and right recursive calls [OK]

Common Mistakes:

• Adding root value before children
• Swapping left and right subtree calls
• Forgetting null check before recursion