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Binary Tree Traversal Interview

Master binary tree traversal coding interviews with our AI-powered coach. Practice inorder, preorder, postorder, and level-order traversals with detailed solutions, complexity analysis, and real-time guidance.

Binary Tree Traversal Interview in Action

Problem: "Implement all four tree traversal methods for a binary tree"

Interviewer: "Given a binary tree, implement inorder, preorder, postorder, and level-order traversals. Explain the time and space complexity of each approach."

Sample Binary Tree

                    1
                   / \
                  2   3
                 / \   \
                4   5   6
               /
              7

Tree Structure:
- Root: 1
- Left subtree: 2 (with children 4, 5)  
- Right subtree: 3 (with child 6)
- Node 4 has left child 7
- Leaf nodes: 7, 5, 6

Expected Traversal Results:
┌─────────────┬─────────────────┬─────────────────────────┐
│ Traversal   │ Order           │ Result                  │
├─────────────┼─────────────────┼─────────────────────────┤
│ Inorder     │ Left-Root-Right │ [7, 4, 2, 5, 1, 3, 6]  │
│ Preorder    │ Root-Left-Right │ [1, 2, 4, 7, 5, 3, 6]  │
│ Postorder   │ Left-Right-Root │ [7, 4, 5, 2, 6, 3, 1]  │
│ Level-order │ Level by Level  │ [1, 2, 3, 4, 5, 6, 7]  │
└─────────────┴─────────────────┴─────────────────────────┘

1. Inorder Traversal (Left Root Right)

// TreeNode class definition class TreeNode { constructor(val, left = null, right = null) { this.val = val; this.left = left; this.right = right; } } // Inorder Traversal - Recursive Approach function inorderTraversal(root) { const result = []; function inorder(node) { if (node === null) return; // Traverse left subtree inorder(node.left); // Process current node result.push(node.val); // Traverse right subtree inorder(node.right); } inorder(root); return result; } // Inorder Traversal - Iterative Approach with Stack function inorderTraversalIterative(root) { if (!root) return []; const result = []; const stack = []; let current = root; while (current || stack.length > 0) { // Go to the leftmost node while (current) { stack.push(current); current = current.left; } // Process current node current = stack.pop(); result.push(current.val); // Move to right subtree current = current.right; } return result; }

Inorder Traversal Analysis:

  • Use Case: Returns values in sorted order for Binary Search Trees
  • Pattern: Left Root Right (LNR)
  • Recursive: Natural and elegant, uses implicit call stack
  • Iterative: Uses explicit stack, better for large trees (avoids stack overflow)
  • Memory: Recursive uses O(h) implicit stack, iterative uses O(h) explicit stack

2. Preorder Traversal (Root Left Right)

// Preorder Traversal - Recursive function preorderTraversal(root) { const result = []; function preorder(node) { if (node === null) return; // Process current node FIRST result.push(node.val); // Then traverse left subtree preorder(node.left); // Finally traverse right subtree preorder(node.right); } preorder(root); return result; } // Preorder Traversal - Iterative function preorderTraversalIterative(root) { if (!root) return []; const result = []; const stack = [root]; while (stack.length > 0) { const node = stack.pop(); // Process current node result.push(node.val); // Push right first, then left (stack is LIFO) if (node.right) stack.push(node.right); if (node.left) stack.push(node.left); } return result; }

3. Postorder Traversal (Left Right Root)

// Postorder Traversal - Recursive function postorderTraversal(root) { const result = []; function postorder(node) { if (node === null) return; // Traverse left subtree first postorder(node.left); // Then traverse right subtree postorder(node.right); // Process current node LAST result.push(node.val); } postorder(root); return result; } // Postorder Traversal - Iterative (Two Stacks Method) function postorderTraversalIterative(root) { if (!root) return []; const result = []; const stack1 = [root]; const stack2 = []; // First pass: traverse and collect nodes in reverse postorder while (stack1.length > 0) { const node = stack1.pop(); stack2.push(node); // Push left first, then right if (node.left) stack1.push(node.left); if (node.right) stack1.push(node.right); } // Second pass: pop from stack2 to get postorder while (stack2.length > 0) { result.push(stack2.pop().val); } return result; }

4. Level-Order Traversal (Breadth-First)

// Level-order Traversal using Queue (BFS) function levelOrder(root) { if (!root) return []; const result = []; const queue = [root]; while (queue.length > 0) { const node = queue.shift(); // Dequeue from front result.push(node.val); // Add children to queue (left to right) if (node.left) queue.push(node.left); if (node.right) queue.push(node.right); } return result; } // Level-order by Levels (2D Array) function levelOrderByLevels(root) { if (!root) return []; const result = []; const queue = [root]; while (queue.length > 0) { const levelSize = queue.length; const currentLevel = []; // Process all nodes at current level for (let i = 0; i < levelSize; i++) { const node = queue.shift(); currentLevel.push(node.val); // Add children for next level if (node.left) queue.push(node.left); if (node.right) queue.push(node.right); } result.push(currentLevel); } return result; }

Time & Space Complexity Analysis

Traversal Method Time Complexity Space Complexity Use Cases
Inorder (Recursive) O(n) O(h) - call stack BST sorted output, expression trees
Inorder (Iterative) O(n) O(h) - explicit stack Avoid recursion limits, controlled memory
Preorder O(n) O(h) Tree copying, prefix expressions
Postorder O(n) O(h) Tree deletion, postfix expressions
Level-order O(n) O(w) - max width Level-wise processing, shortest path

Note: n = number of nodes, h = height of tree, w = maximum width of tree
Best case height: O(log n) for balanced tree
Worst case height: O(n) for skewed tree

Advanced Interview Follow-up Questions:

  • "How would you modify inorder traversal for a threaded binary tree?"
  • "Implement Morris traversal for O(1) space complexity"
  • "How would you do reverse level-order traversal?"
  • "Implement traversal that stops when finding a specific value"
  • "How would you traverse a binary tree in zigzag order?"
  • "What if the tree has parent pointers? How does that change traversal?"

Key Interviewer Expectations:

  • Understand when to use recursive vs iterative approaches
  • Explain space complexity trade-offs
  • Handle edge cases (empty tree, single node)
  • Know practical applications of each traversal type
  • Be able to modify algorithms for specific requirements

🌳 All Traversal Types

Master inorder, preorder, postorder, and level-order traversals with both recursive and iterative implementations.

⚡ Time & Space Analysis

Understand complexity trade-offs, when to use each approach, and how tree shape affects performance.

🔄 Multiple Implementations

Learn both recursive and iterative solutions, understand stack vs queue usage, and handle edge cases.

🎯 Real Applications

Understand practical use cases: BST sorting, expression evaluation, tree copying, and level-wise processing.

🧠 Advanced Patterns

Practice Morris traversal, threaded trees, zigzag traversal, and other advanced tree traversal techniques.

💡 Interview Strategy

Learn how to approach tree problems, identify patterns, and explain your thought process clearly to interviewers.

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