Invert Binary Tree

🎯

Invert Binary Tree

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Imagine you have a family tree diagram and want to see it flipped horizontally, like looking at it in a mirror. This problem asks you to invert a binary tree in a similar way.

💡 This problem is about transforming a binary tree by swapping its left and right children at every node. Beginners often struggle because they may not realize that a simple recursive traversal with swapping can solve it efficiently, and might try complicated or iterative methods first.

📋

Problem Statement

Given the root of a binary tree, invert the tree, and return its root. Inverting a binary tree means swapping the left and right children of all nodes in the tree.

The number of nodes in the tree is in the range [1, 10^5].Node values can be any integer.The tree can be unbalanced.

💡

Example

Input"root = [4,2,7,1,3,6,9]"

Output[4,7,2,9,6,3,1]

The original tree has root 4 with left child 2 and right child 7. After inversion, 2 and 7 are swapped, and this swapping continues recursively for all nodes.

• Single node tree → output is the same single node
• Tree with only left children → output is a tree with only right children
• Tree with only right children → output is a tree with only left children
• Empty tree (null root) → output is null

⚠️

Common Mistakes

Not swapping children after recursive calls

Tree remains unchanged or partially inverted

✅ Ensure swapping left and right children after recursive inversion of subtrees

Swapping children before recursive calls

Leads to incorrect inversion because subtrees are not inverted properly

✅ Perform recursive inversion first, then swap children

Not handling null root (empty tree)

Code may throw null pointer exceptions or runtime errors

✅ Add base case to return null if root is null

Using iterative approach without checking for null children before enqueueing

Queue may contain nulls causing errors during processing

✅ Check if child nodes are not null before adding to queue

Confusing left and right pointers during swap

Tree structure becomes corrupted or inversion is incorrect

✅ Use a temporary variable or tuple unpacking to swap correctly

🧠

Brute Force (Pure Recursion)

💡 Starting with a simple recursive approach helps understand the problem deeply by directly applying the definition of inversion without worrying about optimization.

Intuition

At each node, recursively invert the left and right subtrees, then swap them. This directly follows the problem statement.

Algorithm

1. If the current node is null, return null.
2. Recursively invert the left subtree.
3. Recursively invert the right subtree.
4. Swap the left and right children of the current node.
5. Return the current node.

💡 The recursion naturally traverses the tree postorder, ensuring children are inverted before swapping, which is key to correct inversion.

</>

Code

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def invertTree(root):
    if root is None:
        return None
    left_inverted = invertTree(root.left)
    right_inverted = invertTree(root.right)
    root.left, root.right = right_inverted, left_inverted
    return root

# Example usage:
# root = TreeNode(4, TreeNode(2, TreeNode(1), TreeNode(3)), TreeNode(7, TreeNode(6), TreeNode(9)))
# inverted_root = invertTree(root)

Line Notes

def invertTree(root):Defines the main function to invert the binary tree starting at root

if root is None:Base case: if node is null, nothing to invert, return None

left_inverted = invertTree(root.left)Recursively invert the left subtree first

right_inverted = invertTree(root.right)Recursively invert the right subtree next

root.left, root.right = right_inverted, left_invertedSwap the inverted left and right subtrees to invert current node

return rootReturn the current node after inversion to link back to parent

public class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int val) { this.val = val; }
}

public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        TreeNode leftInverted = invertTree(root.left);
        TreeNode rightInverted = invertTree(root.right);
        root.left = rightInverted;
        root.right = leftInverted;
        return root;
    }

    // Example usage:
    // public static void main(String[] args) {
    //     TreeNode root = new TreeNode(4);
    //     root.left = new TreeNode(2);
    //     root.right = new TreeNode(7);
    //     root.left.left = new TreeNode(1);
    //     root.left.right = new TreeNode(3);
    //     root.right.left = new TreeNode(6);
    //     root.right.right = new TreeNode(9);
    //     Solution sol = new Solution();
    //     TreeNode inverted = sol.invertTree(root);
    // }
}

Line Notes

public TreeNode invertTree(TreeNode root) {Main method to invert the tree starting at root

if (root == null) return null;Base case: if node is null, return null immediately

TreeNode leftInverted = invertTree(root.left);Recursively invert left subtree

TreeNode rightInverted = invertTree(root.right);Recursively invert right subtree

root.left = rightInverted;Assign inverted right subtree to left child

root.right = leftInverted;Assign inverted left subtree to right child

return root;Return current node after inversion

#include <iostream>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

TreeNode* invertTree(TreeNode* root) {
    if (root == nullptr) return nullptr;
    TreeNode* leftInverted = invertTree(root->left);
    TreeNode* rightInverted = invertTree(root->right);
    root->left = rightInverted;
    root->right = leftInverted;
    return root;
}

// Example usage:
// int main() {
//     TreeNode* root = new TreeNode(4);
//     root->left = new TreeNode(2);
//     root->right = new TreeNode(7);
//     root->left->left = new TreeNode(1);
//     root->left->right = new TreeNode(3);
//     root->right->left = new TreeNode(6);
//     root->right->right = new TreeNode(9);
//     TreeNode* inverted = invertTree(root);
//     return 0;
// }

Line Notes

TreeNode* invertTree(TreeNode* root) {Function to invert the binary tree starting at root pointer

if (root == nullptr) return nullptr;Base case: if node pointer is null, return null

TreeNode* leftInverted = invertTree(root->left);Recursively invert left subtree

TreeNode* rightInverted = invertTree(root->right);Recursively invert right subtree

root->left = rightInverted;Assign inverted right subtree to left child pointer

root->right = leftInverted;Assign inverted left subtree to right child pointer

return root;Return current node pointer after inversion

function TreeNode(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
}

function invertTree(root) {
    if (root === null) return null;
    const leftInverted = invertTree(root.left);
    const rightInverted = invertTree(root.right);
    root.left = rightInverted;
    root.right = leftInverted;
    return root;
}

// Example usage:
// const root = new TreeNode(4,
//     new TreeNode(2, new TreeNode(1), new TreeNode(3)),
//     new TreeNode(7, new TreeNode(6), new TreeNode(9))
// );
// const inverted = invertTree(root);
// console.log(inverted);

Line Notes

function invertTree(root) {Defines the function to invert the binary tree starting at root

if (root === null) return null;Base case: if node is null, return null immediately

const leftInverted = invertTree(root.left);Recursively invert left subtree

const rightInverted = invertTree(root.right);Recursively invert right subtree

root.left = rightInverted;Assign inverted right subtree to left child

root.right = leftInverted;Assign inverted left subtree to right child

return root;Return current node after inversion

⏱

Complexity

TimeO(n)

SpaceO(h)

Each node is visited once, so time is O(n). The recursion stack can go as deep as the tree height h, so space is O(h).

💡 For a tree with 1000 nodes and height 10, this means about 1000 operations and stack depth of 10, which is efficient.

Interview Verdict: Accepted

This approach is efficient and accepted for all valid inputs, making it a solid baseline solution.

🧠

Iterative Approach Using Queue (BFS)

💡 This approach uses a queue to invert the tree level by level, which helps understand iterative tree traversal and avoids recursion stack issues.

Intuition

Use a queue to traverse the tree in breadth-first order, swapping children at each node as you go.

Algorithm

1. If the root is null, return null.
2. Initialize a queue and enqueue the root node.
3. While the queue is not empty:
4. Dequeue a node from the queue.
5. Swap its left and right children.
6. If left child is not null, enqueue it.
7. If right child is not null, enqueue it.
8. Return the root node.

💡 This approach uses level order traversal to visit nodes systematically and invert them without recursion.

</>

Code

from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def invertTree(root):
    if root is None:
        return None
    queue = deque([root])
    while queue:
        node = queue.popleft()
        node.left, node.right = node.right, node.left
        if node.left:
            queue.append(node.left)
        if node.right:
            queue.append(node.right)
    return root

# Example usage:
# root = TreeNode(4, TreeNode(2, TreeNode(1), TreeNode(3)), TreeNode(7, TreeNode(6), TreeNode(9)))
# inverted_root = invertTree(root)

Line Notes

from collections import dequeImport deque for efficient queue operations

if root is None:Handle empty tree edge case

queue = deque([root])Initialize queue with root node to start BFS

while queue:Process nodes until queue is empty

node = queue.popleft()Dequeue current node to process

node.left, node.right = node.right, node.leftSwap left and right children of current node

if node.left:If left child exists after swap, enqueue it for later processing

if node.right:If right child exists after swap, enqueue it for later processing

return rootReturn the root of the inverted tree

import java.util.LinkedList;
import java.util.Queue;

public class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int val) { this.val = val; }
}

public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            TreeNode temp = node.left;
            node.left = node.right;
            node.right = temp;
            if (node.left != null) queue.add(node.left);
            if (node.right != null) queue.add(node.right);
        }
        return root;
    }

    // Example usage:
    // public static void main(String[] args) {
    //     TreeNode root = new TreeNode(4);
    //     root.left = new TreeNode(2);
    //     root.right = new TreeNode(7);
    //     root.left.left = new TreeNode(1);
    //     root.left.right = new TreeNode(3);
    //     root.right.left = new TreeNode(6);
    //     root.right.right = new TreeNode(9);
    //     Solution sol = new Solution();
    //     TreeNode inverted = sol.invertTree(root);
    // }
}

Line Notes

if (root == null) return null;Handle empty tree case immediately

Queue<TreeNode> queue = new LinkedList<>();Initialize queue for BFS traversal

queue.add(root);Add root node to start BFS

while (!queue.isEmpty()) {Process nodes until queue is empty

TreeNode node = queue.poll();Dequeue current node

TreeNode temp = node.left;Temporarily store left child for swapping

node.left = node.right;Assign right child to left

node.right = temp;Assign stored left child to right

if (node.left != null) queue.add(node.left);Enqueue left child if exists

if (node.right != null) queue.add(node.right);Enqueue right child if exists

return root;Return root of inverted tree

#include <iostream>
#include <queue>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

TreeNode* invertTree(TreeNode* root) {
    if (root == nullptr) return nullptr;
    queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        TreeNode* node = q.front();
        q.pop();
        TreeNode* temp = node->left;
        node->left = node->right;
        node->right = temp;
        if (node->left) q.push(node->left);
        if (node->right) q.push(node->right);
    }
    return root;
}

// Example usage:
// int main() {
//     TreeNode* root = new TreeNode(4);
//     root->left = new TreeNode(2);
//     root->right = new TreeNode(7);
//     root->left->left = new TreeNode(1);
//     root->left->right = new TreeNode(3);
//     root->right->left = new TreeNode(6);
//     root->right->right = new TreeNode(9);
//     TreeNode* inverted = invertTree(root);
//     return 0;
// }

Line Notes

if (root == nullptr) return nullptr;Handle empty tree case

queue<TreeNode*> q;Initialize queue for BFS

q.push(root);Start BFS with root node

while (!q.empty()) {Process nodes until queue is empty

TreeNode* node = q.front();Get current node from queue front

q.pop();Remove current node from queue

TreeNode* temp = node->left;Store left child temporarily for swapping

node->left = node->right;Assign right child to left

node->right = temp;Assign stored left child to right

if (node->left) q.push(node->left);Enqueue left child if exists

if (node->right) q.push(node->right);Enqueue right child if exists

return root;Return root of inverted tree

function TreeNode(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
}

function invertTree(root) {
    if (root === null) return null;
    const queue = [root];
    while (queue.length > 0) {
        const node = queue.shift();
        [node.left, node.right] = [node.right, node.left];
        if (node.left !== null) queue.push(node.left);
        if (node.right !== null) queue.push(node.right);
    }
    return root;
}

// Example usage:
// const root = new TreeNode(4,
//     new TreeNode(2, new TreeNode(1), new TreeNode(3)),
//     new TreeNode(7, new TreeNode(6), new TreeNode(9))
// );
// const inverted = invertTree(root);
// console.log(inverted);

Line Notes

if (root === null) return null;Handle empty tree edge case

const queue = [root];Initialize queue with root node for BFS

while (queue.length > 0) {Process nodes until queue is empty

const node = queue.shift();Dequeue node from front of queue

[node.left, node.right] = [node.right, node.left];Swap left and right children using destructuring

if (node.left !== null) queue.push(node.left);Enqueue left child if exists

if (node.right !== null) queue.push(node.right);Enqueue right child if exists

return root;Return root of inverted tree

⏱

Complexity

TimeO(n)

SpaceO(n)

Each node is visited once, so time is O(n). The queue can hold up to O(n) nodes in the worst case (e.g., a full level).

💡 For a tree with 1000 nodes, this means about 1000 operations and queue size up to 1000 in worst case, which is manageable.

Interview Verdict: Accepted

This iterative approach is also accepted and useful when recursion depth is a concern.

🧠

Optimized Recursive Approach (In-place Postorder Traversal)

💡 This approach is a refined recursive method that emphasizes in-place swapping during postorder traversal, minimizing extra variables and improving clarity.

Intuition

Invert left and right subtrees recursively, then swap them directly on the current node without extra variables.

Algorithm

1. If the current node is null, return null.
2. Recursively invert the left subtree.
3. Recursively invert the right subtree.
4. Swap the left and right children of the current node in-place.
5. Return the current node.

💡 This approach uses postorder traversal to ensure children are inverted before swapping, which is essential for correctness.

</>

Code

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def invertTree(root):
    if root is None:
        return None
    invertTree(root.left)
    invertTree(root.right)
    root.left, root.right = root.right, root.left
    return root

# Example usage:
# root = TreeNode(4, TreeNode(2, TreeNode(1), TreeNode(3)), TreeNode(7, TreeNode(6), TreeNode(9)))
# inverted_root = invertTree(root)

Line Notes

def invertTree(root):Defines the function to invert the tree starting at root

if root is None:Base case: return None if node is null

invertTree(root.left)Recursively invert left subtree first

invertTree(root.right)Recursively invert right subtree next

root.left, root.right = root.right, root.leftSwap left and right children in-place after inversion

return rootReturn current node after inversion

public class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int val) { this.val = val; }
}

public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        invertTree(root.left);
        invertTree(root.right);
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        return root;
    }

    // Example usage:
    // public static void main(String[] args) {
    //     TreeNode root = new TreeNode(4);
    //     root.left = new TreeNode(2);
    //     root.right = new TreeNode(7);
    //     root.left.left = new TreeNode(1);
    //     root.left.right = new TreeNode(3);
    //     root.right.left = new TreeNode(6);
    //     root.right.right = new TreeNode(9);
    //     Solution sol = new Solution();
    //     TreeNode inverted = sol.invertTree(root);
    // }
}

Line Notes

public TreeNode invertTree(TreeNode root) {Main method to invert tree starting at root

if (root == null) return null;Base case: return null if node is null

invertTree(root.left);Recursively invert left subtree

invertTree(root.right);Recursively invert right subtree

TreeNode temp = root.left;Temporarily store left child for swapping

root.left = root.right;Assign right child to left

root.right = temp;Assign stored left child to right

return root;Return current node after inversion

#include <iostream>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

TreeNode* invertTree(TreeNode* root) {
    if (root == nullptr) return nullptr;
    invertTree(root->left);
    invertTree(root->right);
    TreeNode* temp = root->left;
    root->left = root->right;
    root->right = temp;
    return root;
}

// Example usage:
// int main() {
//     TreeNode* root = new TreeNode(4);
//     root->left = new TreeNode(2);
//     root->right = new TreeNode(7);
//     root->left->left = new TreeNode(1);
//     root->left->right = new TreeNode(3);
//     root->right->left = new TreeNode(6);
//     root->right->right = new TreeNode(9);
//     TreeNode* inverted = invertTree(root);
//     return 0;
// }

Line Notes

TreeNode* invertTree(TreeNode* root) {Function to invert tree starting at root pointer

if (root == nullptr) return nullptr;Base case: return null if node pointer is null

invertTree(root->left);Recursively invert left subtree

invertTree(root->right);Recursively invert right subtree

TreeNode* temp = root->left;Temporarily store left child pointer for swapping

root->left = root->right;Assign right child pointer to left

root->right = temp;Assign stored left child pointer to right

return root;Return current node pointer after inversion

function TreeNode(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
}

function invertTree(root) {
    if (root === null) return null;
    invertTree(root.left);
    invertTree(root.right);
    [root.left, root.right] = [root.right, root.left];
    return root;
}

// Example usage:
// const root = new TreeNode(4,
//     new TreeNode(2, new TreeNode(1), new TreeNode(3)),
//     new TreeNode(7, new TreeNode(6), new TreeNode(9))
// );
// const inverted = invertTree(root);
// console.log(inverted);

Line Notes

function invertTree(root) {Defines function to invert tree starting at root

if (root === null) return null;Base case: return null if node is null

invertTree(root.left);Recursively invert left subtree

invertTree(root.right);Recursively invert right subtree

[root.left, root.right] = [root.right, root.left];Swap left and right children in-place using destructuring

return root;Return current node after inversion

⏱

Complexity

TimeO(n)

SpaceO(h)

Each node is visited once, so time is O(n). The recursion stack depth is O(h), where h is tree height.

💡 For a balanced tree with 1000 nodes, this means about 1000 operations and stack depth about 10, which is efficient.

Interview Verdict: Accepted

This is the cleanest recursive approach and is accepted in interviews and production code.

📊

All Approaches - One-Glance Tradeoffs

💡 In most interviews, the recursive approach (Approach 3) is preferred for its clarity and simplicity. Iterative approach is useful if recursion depth is a concern.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force (Pure Recursion)	O(n)	O(h)	Yes (deep recursion)	N/A	Mention and code as baseline solution
2. Iterative Approach Using Queue (BFS)	O(n)	O(n)	No	N/A	Mention as alternative to recursion
3. Optimized Recursive Approach (In-place Postorder Traversal)	O(n)	O(h)	Yes (deep recursion)	N/A	Preferred approach to code in interviews

💼

Interview Strategy

💡 Use this guide to understand the problem deeply before your interview. Start by clarifying the problem, then explain the brute force recursive approach, followed by iterative and optimized recursive methods. Practice coding and testing each approach.

How to Present

Step 1: Clarify the problem and constraints with the interviewer.Step 2: Describe the brute force recursive approach and its intuition.Step 3: Code the recursive solution and test with examples.Step 4: Discuss iterative BFS approach to avoid recursion stack issues.Step 5: Present the optimized recursive approach for clarity and in-place inversion.Step 6: Analyze time and space complexity for each approach.Step 7: Discuss edge cases and how your code handles them.

Time Allocation

Clarify: 2min → Approach: 3min → Code: 8min → Test: 5min. Total ~18min

What the Interviewer Tests

The interviewer tests your understanding of tree traversal, recursion, and ability to transform trees in-place. They also check if you handle edge cases and can optimize your solution.

Common Follow-ups

• Can you invert the tree iteratively? → Use BFS with a queue to swap nodes level by level.
• What is the space complexity of your recursive solution? → O(h) due to recursion stack, where h is tree height.

💡 These follow-ups test your flexibility with iterative solutions and understanding of recursion space costs.

🔍

Pattern Recognition

When to Use

1) Problem involves binary tree transformation, 2) Requires swapping or rearranging children, 3) Recursive or DFS traversal is natural, 4) Output is a modified tree structure.

Signature Phrases

invert the treeswap left and right childrenmirror image of the tree

NOT This Pattern When

Problems that only require traversal without modification, or those that require path sums or subtree counts.

Similar Problems

Symmetric Tree - checks if a tree is a mirror of itselfMaximum Depth of Binary Tree - uses DFS traversalBinary Tree Upside Down - another tree transformation problem

Practice

1. You are given a binary tree that is complete (all levels except possibly the last are fully filled, and nodes in the last level are as far left as possible). You need to count the total number of nodes efficiently. Which approach guarantees the optimal time complexity for this problem?

easy

A. Perform a simple DFS traversal counting all nodes, which takes O(n) time.

B. Use the tree height to check if subtrees are perfect and recursively count nodes, achieving O((log n)^2) time.

C. Apply a greedy approach by counting nodes level by level until the last incomplete level.

D. Use a breadth-first search (BFS) traversal to count nodes, which is efficient for complete trees.

Solution

1. Step 1: Understand the problem constraints

   The tree is complete, so the height can be used to identify perfect subtrees quickly.

2. Step 2: Recognize the optimal approach

   Using the height to check if left or right subtree is perfect allows skipping large parts of the tree, reducing complexity to O((log n)^2).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Optimal approach leverages tree height and completeness [OK]

Hint: Use tree height to skip perfect subtrees [OK]

Common Mistakes:

• Assuming BFS or simple DFS is optimal
• Using greedy level counting without height checks

2. What is the time complexity of the optimal countNodes algorithm that uses binary search on the last level of a complete binary tree with n nodes?

medium

A. O(log n)

B. O(n)

C. O((log n)^2)

D. O(n log n)

Solution

1. Step 1: Analyze height and binary search

   Height h = O(log n). Binary search on last level does O(log n) steps.

2. Step 2: Combine existence checks

   Each existence check traverses height h = O(log n). Total time = O(log n) * O(log n) = O((log n)^2).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Nested log factors from height and binary search [OK]

Hint: Two nested log factors multiply to (log n)^2 [OK]

Common Mistakes:

• Confusing O(log n) with O((log n)^2)
• Assuming linear time due to traversal

3. Suppose the problem is modified so that cameras can monitor nodes up to distance 2 away (i.e., parent, self, children, and grandchildren). Which change to the algorithm is necessary to maintain minimal camera placement?

hard

A. Use the same three-state postorder traversal but add a fourth state to track grandchildren coverage

B. Place cameras greedily on all nodes with uncovered children as before, no change needed

C. Modify DFS to track coverage states up to distance 2 and place cameras accordingly, increasing state complexity

D. Switch to brute force recursion since greedy no longer works

Solution

1. Step 1: Understand coverage radius increase

   Cameras now cover nodes up to distance 2, so coverage states must reflect grandchildren coverage, not just immediate children.

2. Step 2: Adjust algorithm states

   The original three states are insufficient; DFS must track coverage at a larger radius, increasing complexity and requiring more nuanced state management.

3. Step 3: Evaluate options

   Simply adding a fourth state (A) is vague and insufficient without redesign. Greedy placement without state changes (B) fails minimality. Brute force (D) is inefficient and unnecessary.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Expanded coverage radius requires more complex state tracking [OK]

Hint: Larger coverage radius -> more complex states in DFS [OK]

Common Mistakes:

• Assuming original states suffice for extended coverage
• Relying on naive greedy placement
• Switching to brute force unnecessarily

4. Suppose the problem is modified so that the path can reuse nodes multiple times (i.e., cycles allowed), and node values can be negative. Which of the following approaches correctly adapts the maximum path sum algorithm to handle this variant?

hard

A. Use the same postorder traversal with global max tracking, ignoring negative gains as before.

B. Detect cycles and use dynamic programming with memoization to avoid infinite loops, allowing repeated nodes in paths.

C. Convert the tree to a graph and run Bellman-Ford algorithm to find the maximum sum path allowing cycles.

D. Use depth-first search with pruning to explore all paths, but limit path length to avoid infinite loops.

Solution

1. Step 1: Understand the impact of allowing node reuse

   Allowing cycles means paths can revisit nodes infinitely, so naive DFS or postorder traversal fails due to infinite loops.

2. Step 2: Identify correct approach

   Dynamic programming with cycle detection and memoization prevents infinite recursion and correctly computes max sums with repeated nodes.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Memoization with cycle detection handles repeated nodes safely [OK]

Hint: Cycle detection and memoization needed for repeated nodes [OK]

Common Mistakes:

• Using same postorder traversal causes infinite loops
• Bellman-Ford is for shortest paths, not max sums with negative cycles
• DFS without cycle checks leads to infinite recursion

5. Suppose you want to extend the serialization/deserialization to support binary trees where nodes can have duplicate values and the tree can be very deep (height > 10,000). Which modification is necessary to ensure correctness and efficiency?

hard

A. Use iterative BFS serialization with null markers and iterative deserialization to avoid recursion stack overflow.

B. Use recursive DFS with memoization to handle duplicates and deep trees efficiently.

C. Switch to preorder traversal without null markers to reduce string size and recursion depth.

D. Serialize only unique node values and reconstruct tree assuming balanced shape.

Solution

1. Step 1: Identify problem with deep recursion

   Recursive DFS can cause stack overflow on very deep trees.

2. Step 2: Use iterative BFS with null markers

   Iterative BFS avoids recursion stack issues and null markers preserve structure even with duplicates.

3. Step 3: Avoid assumptions about uniqueness or balanced shape

   Duplicates require storing all nodes explicitly; balanced assumptions break correctness.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Iterative BFS with null markers handles deep trees and duplicates safely [OK]

Hint: Iterative BFS avoids recursion limits and preserves structure [OK]

Common Mistakes:

• Removing null markers to save space breaks reconstruction
• Using recursion on deep trees causes stack overflow
• Assuming unique values or balanced trees