Tree Traversals - Learning Module

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Use Cases for Each Traversal Order

Choosing the Right Traversal

You've now learned four tree traversals: preorder, inorder, postorder (all DFS-style), and level-order (BFS). Each visits every node exactly once, yet each produces a different order with different properties. The question that separates beginners from experts is: given a specific problem, which traversal should I use?

This isn't about memorizing a lookup table. It's about understanding the fundamental properties of each traversal and matching them to problem requirements. By the end of this page, you'll have a systematic framework for choosing the right traversal every time.

What You Will Learn

By the end of this page, you will have a complete decision framework for selecting tree traversals. You'll understand the key question to ask, see comprehensive use case mappings, and practice applying the framework to real problems. This is the synthesis page that ties all traversal knowledge together.

The Complete Traversal Reference

Let's begin with a comprehensive reference comparing all four traversals side by side:

all-traversals.txt
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THE FOUR TREE TRAVERSALS
═══════════════════════════════════════════════════════════════════════════
 
Sample Tree:
                    A
                   / \
                  B   C
                 / \   \
                D   E   F
 
═══════════════════════════════════════════════════════════════════════════
 
Traversal      │ Order                 │ Output        │ Visit Position
───────────────┼───────────────────────┼───────────────┼─────────────────────
Preorder       │ Root → Left → Right   │ A,B,D,E,C,F   │ Before children
Inorder        │ Left → Root → Right   │ D,B,E,A,C,F   │ Between children
Postorder      │ Left → Right → Root   │ D,E,B,F,C,A   │ After children
Level-order    │ Level by level        │ A,B,C,D,E,F   │ Top to bottom
 
═══════════════════════════════════════════════════════════════════════════
 
Key Observations:
• Preorder:   Root (A) is FIRST
• Postorder:  Root (A) is LAST
• Inorder:    Root (A) is in the MIDDLE (for this balanced tree)
• Level-order: Nodes appear in top-to-bottom, left-to-right order
 
═══════════════════════════════════════════════════════════════════════════
 
Data Structure Used:
• Preorder:   Stack (explicit or implicit via recursion)
• Inorder:    Stack (explicit or implicit via recursion)
• Postorder:  Stack (explicit or implicit via recursion)
• Level-order: Queue

Traversal Properties at a Glance
Property	Preorder	Inorder	Postorder	Level-order
Order	VLR	LVR	LRV	Level by level
Root position	First	Middle*	Last	First
Approach	DFS	DFS	DFS	BFS
Data structure	Stack	Stack	Stack	Queue
Time	O(n)	O(n)	O(n)	O(n)
Space	O(h)	O(h)	O(h)	O(w)
Information flow	Top-down	Mixed	Bottom-up	Top-down by level

Space Complexity Note

h = tree height, w = maximum tree width. For a balanced tree: h ≈ log n and w ≈ n/2. For a skewed tree: h ≈ n and w ≈ 1. This means DFS is more space-efficient for wide, shallow trees while BFS is better for narrow, deep trees.

The Decision Framework: Three Key Questions

When facing a tree problem, ask yourself these three questions in order. The answers will guide you to the correct traversal:

Question 1: Does the problem require level-by-level processing?

•Do you need to process nodes level by level?
•Is the problem about depth/level of nodes?
•Do you need the shortest path (in terms of edges)?
•Are you computing level averages, sums, or comparisons?
•If YES → Use Level-order (BFS)

Question 2: What is the information flow direction?

•Top-down: Does parent provide info to children? → Preorder
•Bottom-up: Do children provide info to parent? → Postorder
•Examples of top-down: passing depth, constraints, accumulated path
•Examples of bottom-up: computing height, size, aggregate values

Question 3: Is this a BST with ordering requirements?

•Do you need elements in sorted order?
•Are you validating BST property?
•Are you finding k-th smallest/largest?
•If YES → Use Inorder

decision-flowchart.txt
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TRAVERSAL DECISION FLOWCHART
═══════════════════════════════════════════════════════════════════════════
 
           ┌─────────────────────────────────────────┐
           │  Need level-by-level processing?        │
           └──────────────────┬──────────────────────┘
                              │
           ┌──────────────────┼──────────────────┐
           │ YES              │                  │ NO
           ▼                  │                  ▼
    ┌──────────────┐          │        ┌─────────────────────────────┐
    │ LEVEL-ORDER  │          │        │ Is this a BST needing      │
    │ (BFS)        │          │        │ sorted access or           │
    └──────────────┘          │        │ ordering validation?       │
                              │        └──────────────┬──────────────┘
                              │                       │
                              │        ┌──────────────┼──────────────┐
                              │        │ YES          │              │ NO
                              │        ▼              │              ▼
                              │ ┌──────────────┐      │   ┌────────────────────────┐
                              │ │   INORDER    │      │   │ Does processing        │
                              │ └──────────────┘      │   │ current node DEPEND    │
                              │                       │   │ on children's results? │
                              │                       │   └───────────┬────────────┘
                              │                       │               │
                              │                       │ ┌─────────────┼─────────────┐
                              │                       │ │ YES         │             │ NO
                              │                       │ ▼             │             ▼
                              │                       │ ┌───────────┐ │  ┌───────────┐
                              │                       │ │ POSTORDER │ │  │ PREORDER  │
                              │                       │ └───────────┘ │  └───────────┘
                              │                       │               │
                              └───────────────────────┴───────────────┘

Preorder Use Cases: "Process Before Descending"

Preorder (Root → Left → Right) is the right choice when you need to process a node before processing its children. Think of it as "do something, then delegate to children."

The defining characteristic: Top-down information flow. Parent's information is needed by children.

Comprehensive Preorder Use Cases
Use Case	Why Preorder	Example
Cloning/copying a tree	Must create parent before attaching children	Deep copy of data structure
Serializing a tree	Write structure (parent) before content (children)	JSON serialization, file format
Printing tree structure	Print parent, then indent children	File system tree, org chart
Prefix expression	Operator before operands	Polish notation
Creating a copy with modification	Modify while copying top-down	Transform tree during traversal
Path problems (root to any node)	Build path as you descend	Find all root-to-leaf paths
Passing constraints down	Parent establishes constraints for children	BST validation with min/max
Computing depth/level	Pass depth + 1 to children	Find depth of target node

preorder-patterns.ts
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// PATTERN 1: Need to do something BEFORE recursing
function process(node, info) {
    doSomething(node, info);      // ← Process first
    process(node.left, newInfo);   // Then recurse
    process(node.right, newInfo);
}
 
// PATTERN 2: Passing information DOWN the tree
function topDown(node, accumulatedValue) {
    // Use accumulated value at current node
    useValue(node, accumulatedValue);
    // Pass updated value to children
    topDown(node.left, update(accumulatedValue));
    topDown(node.right, update(accumulatedValue));
}
 
// PATTERN 3: Creating/building something
function create(node) {
    const newNode = { value: transform(node.value) };  // CREATE first
    newNode.left = create(node.left);   // Then create children
    newNode.right = create(node.right);
    return newNode;
}
 
// CONCRETE EXAMPLE: Print paths from root
function printPaths(node, path = []) {
    if (!node) return;
    
    path.push(node.value);  // PREORDER: add to path first
    
    if (!node.left && !node.right) {
        console.log(path.join(' → '));
    }
    
    printPaths(node.left, [...path]);
    printPaths(node.right, [...path]);
}

Inorder Use Cases: "BST and Sorted Access"

Inorder (Left → Root → Right) is primarily associated with Binary Search Trees. For BSTs, inorder traversal produces elements in sorted order.

The defining characteristic: For BSTs, inorder visits nodes in ascending value order. For expression trees, inorder produces infix notation.

Comprehensive Inorder Use Cases
Use Case	Why Inorder	Example
Sorted iteration of BST	Inorder = sorted order for BST	Print all elements sorted
K-th smallest element	Count in inorder, stop at k	Find median in BST
K-th largest element	Reverse inorder (RVL), stop at k	Find 3rd largest
BST validation	Check each value > previous	Is binary tree a valid BST?
BST to sorted array	Inorder gives sorted result	Flatten BST
Infix expression	Operator between operands	Standard math notation
Range queries in BST	Stop when out of range	Find all values between L and R
BST Iterator	On-demand next smallest	Merge K sorted BSTs

inorder-patterns.ts
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// PATTERN 1: Need sorted order from BST
function getSorted(bstNode) {
    const result = [];
    function inorder(node) {
        if (!node) return;
        inorder(node.left);        // Left first
        result.push(node.value);   // Then this node
        inorder(node.right);       // Then right
    }
    inorder(bstNode);
    return result;  // Sorted!
}
 
// PATTERN 2: Previous value comparison (BST validation)
function isValidBST(node, prev = { value: -Infinity }) {
    if (!node) return true;
    
    if (!isValidBST(node.left, prev)) return false;
    
    // INORDER: check ordering
    if (node.value <= prev.value) return false;
    prev.value = node.value;
    
    return isValidBST(node.right, prev);
}
 
// PATTERN 3: K-th element (counting in order)
function kthSmallest(node, k) {
    let count = 0;
    let result = null;
    
    function inorder(node) {
        if (!node || result !== null) return;
        
        inorder(node.left);
        
        count++;
        if (count === k) {
            result = node.value;
            return;
        }
        
        inorder(node.right);
    }
    
    inorder(node);
    return result;
}
 
// CONCRETE EXAMPLE: Find all nodes in range [low, high]
function rangeSearch(node, low, high, result = []) {
    if (!node) return result;
    
    // Only go left if there might be values >= low
    if (node.value > low) {
        rangeSearch(node.left, low, high, result);
    }
    
    // Include this node if in range
    if (node.value >= low && node.value <= high) {
        result.push(node.value);
    }
    
    // Only go right if there might be values <= high
    if (node.value < high) {
        rangeSearch(node.right, low, high, result);
    }
    
    return result;
}

Inorder Is NOT Just for BSTs

While inorder is most powerful for BSTs, it can be useful for general binary trees too—particularly for expression trees (infix notation) or when you need a symmetric traversal pattern. However, for non-BST trees, inorder doesn't have the magical 'sorted output' property.

Postorder Use Cases: "Children Before Parent"

Postorder (Left → Right → Root) is the right choice when processing a node requires information from its children. Think of it as "wait for children to report, then aggregate."

The defining characteristic: Bottom-up information flow. Children's information is needed by parent.

Comprehensive Postorder Use Cases
Use Case	Why Postorder	Example
Tree deletion	Delete children before parent	Memory deallocation
Computing height	Height = 1 + max(children heights)	Find tree depth
Computing size	Size = 1 + left size + right size	Count all nodes
Computing sum	Sum = value + left sum + right sum	Sum all values
Tree diameter	Need heights to compute path through node	Longest path
Balance check	Compare children heights	Is tree balanced?
Subtree problems	Compute subtree property first	Largest BST subtree
Postfix expression	Operands before operator	RPN calculator
Expression evaluation	Evaluate operands before applying operator	Compute result
Maximum path sum	Compare paths through children	Binary tree max path

postorder-patterns.ts
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// PATTERN 1: Compute from children's results
function computeFromChildren(node) {
    if (!node) return baseCase;
    
    const leftResult = computeFromChildren(node.left);
    const rightResult = computeFromChildren(node.right);
    
    // POSTORDER: use children's results to compute this node's result
    return combine(node.value, leftResult, rightResult);
}
 
// PATTERN 2: Aggregation (sum, max, count, etc.)
function aggregate(node) {
    if (!node) return 0;
    
    const leftVal = aggregate(node.left);
    const rightVal = aggregate(node.right);
    
    return node.value + leftVal + rightVal;  // Current depends on children
}
 
// PATTERN 3: Property checking that needs children's properties
function checkProperty(node) {
    if (!node) return { valid: true, property: 0 };
    
    const leftCheck = checkProperty(node.left);
    const rightCheck = checkProperty(node.right);
    
    // POSTORDER: check this node using children's info
    const thisValid = leftCheck.valid && rightCheck.valid && 
                      someCondition(leftCheck, rightCheck, node);
    
    return { valid: thisValid, property: computeProperty(...) };
}
 
// CONCRETE EXAMPLE: Tree diameter (max path between any two nodes)
function diameter(node) {
    let maxDiameter = 0;
    
    function height(node) {
        if (!node) return -1;
        
        const leftH = height(node.left);
        const rightH = height(node.right);
        
        // POSTORDER: update diameter using children's heights
        maxDiameter = Math.max(maxDiameter, leftH + rightH + 2);
        
        return 1 + Math.max(leftH, rightH);
    }
    
    height(node);
    return maxDiameter;
}

Level-Order Use Cases: "Level by Level"

Level-order (BFS) processes nodes level by level, from top to bottom. This is the only traversal that isn't naturally recursive—it uses a queue instead of a stack.

The defining characteristic: Processes nodes by their distance from the root. All nodes at depth d are processed before any node at depth d+1.

Comprehensive Level-Order Use Cases
Use Case	Why Level-Order	Example
Print tree level by level	Naturally groups by level	Visualize tree structure
Level averages/sums	Process complete levels	Average of levels
Minimum depth to leaf	First leaf found is min depth	Shortest path to leaf
Right/left side view	Last/first of each level	Side view of tree
Connect nodes at same level	Access siblings in queue	Populate next pointers
Zigzag level order	Alternate left/right by level	Zigzag traversal
Check completeness	Detect gaps level by level	Is complete binary tree?
Serialize for breadth structure	Level by level is natural for BFS	BFS-friendly serialization
Maximum width of tree	Count nodes per level	Widest level in tree

level-order-patterns.ts
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// PATTERN 1: Classic BFS with queue
function levelOrder(root) {
    if (!root) return [];
    
    const result = [];
    const queue = [root];
    
    while (queue.length > 0) {
        const node = queue.shift();
        result.push(node.value);
        
        if (node.left) queue.push(node.left);
        if (node.right) queue.push(node.right);
    }
    
    return result;
}
 
// PATTERN 2: Level-by-level processing (grouped)
function levelsByLevel(root) {
    if (!root) return [];
    
    const result = [];
    const queue = [root];
    
    while (queue.length > 0) {
        const levelSize = queue.length;  // Key: snapshot current level size
        const currentLevel = [];
        
        for (let i = 0; i < levelSize; i++) {
            const node = queue.shift();
            currentLevel.push(node.value);
            
            if (node.left) queue.push(node.left);
            if (node.right) queue.push(node.right);
        }
        
        result.push(currentLevel);  // Each level is separate array
    }
    
    return result;
}
 
// PATTERN 3: Early termination (find something on nearest level)
function minDepthToLeaf(root) {
    if (!root) return 0;
    
    const queue = [{ node: root, depth: 1 }];
    
    while (queue.length > 0) {
        const { node, depth } = queue.shift();
        
        // BFS finds shortest path first!
        if (!node.left && !node.right) {
            return depth;  // First leaf = minimum depth
        }
        
        if (node.left) queue.push({ node: node.left, depth: depth + 1 });
        if (node.right) queue.push({ node: node.right, depth: depth + 1 });
    }
}
 
// CONCRETE EXAMPLE: Right side view of tree
function rightSideView(root) {
    if (!root) return [];
    
    const result = [];
    const queue = [root];
    
    while (queue.length > 0) {
        const levelSize = queue.length;
        
        for (let i = 0; i < levelSize; i++) {
            const node = queue.shift();
            
            // Last node at this level = rightmost = visible from right
            if (i === levelSize - 1) {
                result.push(node.value);
            }
            
            if (node.left) queue.push(node.left);
            if (node.right) queue.push(node.right);
        }
    }
    
    return result;
}

Problem → Traversal Quick Reference

Here's a comprehensive mapping from common problem types to the appropriate traversal. Use this as a quick reference when solving tree problems:

Problem to Traversal Mapping
Problem Type	Best Traversal	Reason
Clone/copy tree	Preorder	Create parent before children
Serialize tree	Preorder	Record structure during descent
Print directory structure	Preorder	Parent before contents
Path from root tracking	Preorder	Build path as you descend
Prefix notation	Preorder	Operator before operands
BST to sorted array	Inorder	BST inorder = sorted
K-th smallest in BST	Inorder	Count to k in sorted order
Validate BST	Inorder	Check each > previous
BST range query	Inorder	Prune + collect in order
Infix notation	Inorder	Operator between operands
Delete tree	Postorder	Delete children first
Compute height	Postorder	Need children heights first
Compute size	Postorder	Need children sizes first
Check balanced	Postorder	Compare children heights
Tree diameter	Postorder	Need heights for path calc
Subtree sum/max	Postorder	Aggregate from children
Postfix notation	Postorder	Operands before operator
Level averages	Level-order	Process level by level
Minimum depth	Level-order	BFS finds shortest first
Right/left side view	Level-order	Access last/first in level
Zigzag traversal	Level-order	Control direction per level
Connect same-level nodes	Level-order	Siblings in queue together
Maximum tree width	Level-order	Count nodes per level

When Traversal Order Doesn't Matter

Some problems just need to visit all nodes—the order is irrelevant. In these cases, use whichever traversal you find easiest to implement (usually preorder for its simplicity):

Order-Independent Problems

•Count total nodes — Just count; order doesn't affect count.
•Find if value exists — Search until found; any order works.
•Sum all values — Addition is commutative; order doesn't matter.
•Find maximum/minimum — Comparison over all; order irrelevant.
•Check if all values satisfy condition — Must check all; order irrelevant.
•Collect all values into set — Sets are unordered anyway.

order-independent.ts
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// All four traversals give the same answer for these:
 
// Count nodes
function countNodes(node) {
    if (!node) return 0;
    return 1 + countNodes(node.left) + countNodes(node.right);
}
 
// Sum all values
function sumTree(node) {
    if (!node) return 0;
    return node.value + sumTree(node.left) + sumTree(node.right);
}
 
// Find maximum
function findMax(node) {
    if (!node) return -Infinity;
    return Math.max(node.value, findMax(node.left), findMax(node.right));
}
 
// Check if value exists
function contains(node, target) {
    if (!node) return false;
    if (node.value === target) return true;
    return contains(node.left, target) || contains(node.right, target);
}
 
/*
 * For these problems, I used a pattern that's roughly preorder-ish,
 * but it doesn't really matter. I could swap the order of
 * processing and the answer would be the same.
 * 
 * When you recognize an order-independent problem, just pick
 * the simplest implementation and move on.
 */

Default to Simplicity

When order doesn't matter, preorder is often simplest because you process the current node first, then recurse. It's the most 'natural' recursive pattern for many developers. But don't overthink it—any traversal works, so use what's comfortable.

Practice: Which Traversal?

Let's practice the decision framework. For each problem, identify the correct traversal before looking at the answer:

practice-traversals.txt
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PRACTICE: WHICH TRAVERSAL?
═══════════════════════════════════════════════════════════════════════════
 
Problem 1: Find the lowest common ancestor of two nodes
───────────────────────────────────────────────────────
Think: Do I need children's info to determine if current node is LCA?
Answer: POSTORDER - Need to check if target nodes are in left/right subtrees
 
Problem 2: Mirror (invert) a binary tree
─────────────────────────────────────────
Think: When do I swap children?
Answer: POSTORDER (swap after children are mirrored) OR PREORDER (swap first)
Both work! Preorder is often cleaner.
 
Problem 3: Find all nodes at depth K
────────────────────────────────────
Think: Do I need level-by-level processing?
Answer: LEVEL-ORDER (stop at level K) OR PREORDER (pass depth parameter)
Level-order is cleaner; preorder works too.
 
Problem 4: Convert BST to Greater Sum Tree
──────────────────────────────────────────
(Each node's value = sum of all greater values + its value)
Think: Need values in descending order...
Answer: REVERSE INORDER (Right → Root → Left) - gives descending order
 
Problem 5: Maximum depth of binary tree
───────────────────────────────────────
Think: Depth at current = 1 + max(depth of children)
Answer: POSTORDER - need children's depths first
 
Problem 6: Flatten binary tree to linked list (in preorder order)
─────────────────────────────────────────────────────────────────
Think: Result should be in preorder sequence...
Answer: PREORDER - naturally produces the desired order
 
Problem 7: All paths from root where sum equals target
──────────────────────────────────────────────────────
Think: Building path from root, accumulating sum...
Answer: PREORDER - track path as you descend, check at leaves
 
Problem 8: Number of nodes in each level
────────────────────────────────────────
Think: Need to group by level...
Answer: LEVEL-ORDER - naturally processes level by level
 
═══════════════════════════════════════════════════════════════════════════
 
KEY INSIGHT: The traversal choice follows from understanding the
information flow and dependencies in the problem!

Summary: Mastering Traversal Selection

You now have a complete framework for choosing the right tree traversal. Let's consolidate the key decision factors:

traversal-cheat-sheet.txt
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TREE TRAVERSAL SELECTION CHEAT SHEET
═══════════════════════════════════════════════════════════════════════════
 
USE PREORDER (Root → Left → Right) WHEN:
────────────────────────────────────────
✓ Creating/cloning tree structure
✓ Serializing (parent info comes first)
✓ Passing information DOWN (top-down)
✓ Building paths from root
✓ Prefix notation needed
 
USE INORDER (Left → Root → Right) WHEN:
───────────────────────────────────────
✓ Need BST elements in sorted order
✓ K-th smallest/largest in BST
✓ Validating BST property
✓ Infix notation needed
✓ Any BST ordering operation
 
USE POSTORDER (Left → Right → Root) WHEN:
─────────────────────────────────────────
✓ Computing from children's values (height, size)
✓ Deleting/freeing tree memory
✓ Result depends on subtree computations (bottom-up)
✓ Postfix notation/expression evaluation
 
USE LEVEL-ORDER (BFS) WHEN:
───────────────────────────
✓ Level-by-level processing required
✓ Finding shortest path (in edges)
✓ Computing level stats (averages, widths)
✓ Side views of tree
✓ Connecting same-level nodes
 
WHEN ORDER DOESN'T MATTER:
──────────────────────────
✓ Simple aggregation (count, sum, max)
✓ Existence checks
✓ Default to preorder (simplest)

Key Takeaways

•Think about information flow — Top-down suggests preorder; bottom-up suggests postorder.
•BST + ordering = Inorder — This is almost always the right answer for BST order-related problems.
•Level-by-level = Level-order — If you need to group by depth, use BFS.
•Dependencies determine order — If you need X to compute Y, X must be traversed first.
•Practice pattern recognition — The more problems you solve, the faster you'll recognize which traversal fits.
•Multiple traversals may work — Sometimes more than one traversal solves the problem; choose the cleanest.

Module Complete:

You have now mastered tree traversals—from understanding what traversal means, through implementing all three DFS traversals (preorder, inorder, postorder), to knowing exactly when to use each one. This knowledge forms the foundation for all tree algorithm work. Every tree problem you encounter will use one of these traversals as its backbone.

Module 5 Complete: Tree Traversals Mastered

Congratulations! You now understand tree traversals at an expert level. You can implement preorder, inorder, and postorder both recursively and iteratively. You understand the unique properties of each (sorted BST output for inorder, bottom-up for postorder, top-down for preorder). Most importantly, you have a systematic framework for choosing the right traversal for any problem. This is knowledge that will serve you throughout your programming career.