Populating Next Right Pointers in Each Node

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Problem

Imagine you are designing a communication system for a perfect binary tree where each node needs to instantly know its immediate right neighbor at the same level to pass messages efficiently.

Given a perfect binary tree, populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. Initially, all next pointers are set to NULL. Input: The root of the perfect binary tree. Output: The same tree with each node's next pointer set to its next right node.

The number of nodes in the tree is in the range [1, 10^5].The tree is perfect: all leaves are on the same level, and every parent has two children.

Edge cases: Single node tree → next pointer remains NULLTree with only two levels → only root's children connectedLarge tree with maximum nodes → performance and memory usage tested

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IDE

def connect(root: 'Node') -> 'Node':public Node connect(Node root)Node* connect(Node* root)function connect(root)

def connect(root: 'Node') -> 'Node':
    # Write your solution here
    pass

class Solution {
    public Node connect(Node root) {
        // Write your solution here
        return root;
    }
}

#include <vector>
using namespace std;

Node* connect(Node* root) {
    // Write your solution here
    return root;
}

function connect(root) {
    // Write your solution here
    return root;
}

Common Bugs to Avoid

Wrong: Next pointers only connect siblings, not across parents at the same level.Greedy approach connecting only left and right children of the same parent, missing connections between parents.✅ Traverse each level fully and connect nodes sequentially, including across different parents using BFS or next pointers.

Wrong: Last node at each level has next pointer not set to NULL but pointing to some node.Off-by-one error in loop causing last node's next pointer to be incorrectly assigned.✅ Explicitly set last node's next pointer to NULL after processing each level.

Wrong: Code crashes or returns non-null for empty tree input.No null check for root before processing.✅ Add a check at the start: if root is null, return null immediately.

Wrong: Next pointers not set or partially set in single node tree.Code assumes at least two nodes and does not handle single node case.✅ Handle single node case by ensuring its next pointer is set to NULL or remains unchanged.

Wrong: Solution times out on large input.Using BFS with queue or recursion causing O(n) space and overhead.✅ Implement O(1) space solution using next pointers to traverse levels iteratively.

✓

Test Cases

Try testing your solution on minimal trees like empty or single node to catch base case bugs.

Think about common pitfalls like only connecting siblings and missing cross-parent links.

Focus on optimizing space complexity by using next pointers instead of extra data structures.

t1_01basic

Input{"root":[1,2,3,4,5,6,7]}

Expected"[1 -> NULL, 2 -> 3, 3 -> NULL, 4 -> 5, 5 -> 6, 6 -> 7, 7 -> NULL]"

⏱ Performance - must finish in 2000ms

At level 1, node 1's next is NULL. At level 2, node 2's next is 3, node 3's next is NULL. At level 3, nodes 4,5,6,7 are connected in order with the last node's next as NULL.

💡 Think about connecting nodes level by level using BFS.

💡 Use a queue to process nodes in each level and link them sequentially.

💡 After processing each level, set the last node's next pointer to NULL explicitly.

Why it failed: Failed to connect nodes at each level correctly; likely missed setting last node's next to NULL or linking nodes in order. Fix by ensuring within each level, nodes are linked in sequence and last node's next is NULL.

✓ Correctly linked all nodes level by level with proper next pointers.

t1_02basic

Input{"root":[10,5,15,3,7,12,18]}

Expected"[10 -> NULL, 5 -> 15, 15 -> NULL, 3 -> 7, 7 -> 12, 12 -> 18, 18 -> NULL]"

⏱ Performance - must finish in 2000ms

At level 1, node 10's next is NULL. At level 2, node 5's next is 15, node 15's next is NULL. At level 3, nodes 3,7,12,18 are connected in order with the last node's next as NULL.

💡 Apply the same BFS level order traversal to this different tree.

💡 Ensure each level's nodes are connected left to right, ending with NULL.

💡 Remember to enqueue children nodes for the next level processing.

Why it failed: Incorrect next pointers due to not processing all nodes in a level or incorrect linking order. Fix by processing nodes level-wise and linking them sequentially.

✓ Successfully connected all nodes in this tree with correct next pointers.

t2_01edge

Input{"root":null}

Expectednull

⏱ Performance - must finish in 2000ms

Empty tree input should return null without errors.

💡 Check if the root is null before processing.

💡 Return null immediately if the tree is empty.

💡 Avoid any operations on null to prevent runtime errors.

Why it failed: Code crashes or returns non-null for empty input. Fix by adding a null check at the start and returning null if root is null.

✓ Correctly handles empty tree input by returning null.

t2_02edge

Input{"root":[1]}

Expected"[1 -> NULL]"

⏱ Performance - must finish in 2000ms

Single node tree's next pointer remains NULL.

💡 For a single node, no next right node exists.

💡 Ensure the node's next pointer is set to NULL explicitly or remains unchanged.

💡 No children to enqueue or connect.

Why it failed: Next pointer incorrectly set to non-null or code fails on single node. Fix by handling single node case and setting next to NULL.

✓ Single node's next pointer correctly remains NULL.

t2_03edge

Input{"root":[1,2,3]}

Expected"[1 -> NULL, 2 -> 3, 3 -> NULL]"

⏱ Performance - must finish in 2000ms

Tree with two levels: root's children connected, last node's next is NULL.

💡 Process the root level first, then its children.

💡 Connect left child to right child at second level.

💡 Set the last child's next pointer to NULL.

Why it failed: Children not connected or last node's next pointer not set to NULL. Fix by linking left child to right child and setting last child's next to NULL.

✓ Two-level tree children correctly connected with proper next pointers.

t3_01corner

Input{"root":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]}

Expected"[1 -> NULL, 2 -> 3, 3 -> NULL, 4 -> 5, 5 -> 6, 6 -> 7, 7 -> NULL, 8 -> 9, 9 -> 10, 10 -> 11, 11 -> 12, 12 -> 13, 13 -> 14, 14 -> 15, 15 -> NULL]"

⏱ Performance - must finish in 2000ms

Perfect binary tree with 4 levels; all nodes connected correctly at each level.

💡 Beware of greedy approaches that connect nodes only within the same parent.

💡 Ensure connections span across parents at the same level.

💡 Use BFS or next pointer traversal to connect nodes across different parents.

Why it failed: Greedy approach connecting only siblings but not across parents causes incorrect next pointers. Fix by connecting nodes across parents using BFS or next pointers.

✓ All nodes connected correctly across parents at each level.

t3_02corner

Input{"root":[1,2,3,4,5,6,7]}

Expected"[1 -> NULL, 2 -> 3, 3 -> NULL, 4 -> 5, 5 -> 6, 6 -> 7, 7 -> NULL]"

⏱ Performance - must finish in 2000ms

Test to catch confusion between 0/1 and unbounded knapsack analog: here, each node must be connected exactly once to next right node.

💡 Do not connect nodes multiple times or skip nodes.

💡 Traverse level by level and connect each node to its immediate right neighbor.

💡 Avoid reusing next pointers incorrectly or skipping nodes.

Why it failed: Incorrect connections due to reusing or skipping nodes; next pointers not set exactly once per node. Fix by traversing level once and connecting each node to its immediate right neighbor.

✓ Each node connected exactly once to its next right node.

t3_03corner

Input{"root":[1,2,3,4,5,6,7]}

Expected"[1 -> NULL, 2 -> 3, 3 -> NULL, 4 -> 5, 5 -> 6, 6 -> 7, 7 -> NULL]"

⏱ Performance - must finish in 2000ms

Test to catch off-by-one errors in linking last node's next pointer at each level.

💡 Ensure the last node at each level has next pointer set to NULL.

💡 Do not link last node to any node beyond its level.

💡 Check loop boundaries carefully when linking nodes.

Why it failed: Last node's next pointer incorrectly set to non-null or linked to wrong node. Fix by explicitly setting last node's next to NULL after linking all nodes in level.

✓ Last node at each level correctly has next pointer set to NULL.

t4_01performance

Input{"root":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127]}

Expectednull

⏱ Performance - must finish in 2000ms

Large perfect binary tree with 127 nodes (7 levels) to test O(n) time complexity and O(1) space optimal solution within 2 seconds.

💡 Use O(1) space approach by leveraging next pointers to traverse levels.

💡 Avoid using extra queues or recursion that increase space complexity.

💡 Process each level using already established next pointers to connect next level.

Why it failed: Solution exceeds time limit due to O(n) space BFS or recursion causing overhead. Fix by implementing O(1) space iterative approach using next pointers.

✓ Solution runs within time limit using O(n) time and O(1) space complexity.

Practice

1. You need to find the value of the leftmost node at the deepest level of a binary tree. Which approach guarantees an optimal solution that efficiently finds this value without traversing unnecessary nodes multiple times?

easy

A. Breadth-first search (BFS) level order traversal inserting right child before left child to ensure the last node processed is the bottom-left node.

B. Depth-first search (DFS) that tracks the maximum depth and updates the leftmost node when a deeper level is found.

C. Greedy traversal always choosing the left child first until reaching the deepest node.

D. Dynamic programming storing results of subtrees to combine leftmost values at each depth.

Solution

1. Step 1: Understand problem goal

   The problem requires finding the bottom-left value, which is the leftmost node at the deepest level.

2. Step 2: Analyze BFS approach with right-to-left insertion

   BFS with queue inserting right child before left child ensures the last node popped is the bottom-left node, but this approach requires careful enqueue order and may be less intuitive.

3. Step 3: Analyze DFS approach

   DFS that tracks maximum depth and updates the leftmost node when a deeper level is found directly finds the bottom-left node efficiently without relying on enqueue order.

4. Step 4: Choose optimal approach

   DFS approach is simpler and guarantees correct bottom-left node without tricky enqueue order.

5. Final Answer:

   Option B -> Option B

6. Quick Check:

   DFS tracking max depth and leftmost node is a standard, efficient solution [OK]

Hint: DFS tracking max depth and leftmost node efficiently finds bottom-left node [OK]

Common Mistakes:

• Assuming BFS with right-to-left insertion is always simpler
• Thinking greedy left child traversal always finds bottom-left
• Misapplying DP to this problem

2. You need to find the maximum value at each depth level of a binary tree. Which approach guarantees an efficient and correct solution for this problem?

easy

A. Perform a breadth-first search (BFS) level order traversal, tracking the maximum value per level.

B. Use a greedy traversal that picks the largest child node at each step without level tracking.

C. Apply a dynamic programming approach that stores maximum values for subtrees without level information.

D. Use a depth-first search (DFS) without tracking levels, updating a global maximum value.

Solution

1. Step 1: Understand the problem requires per-level maximums

   The problem asks for the largest value at each tree row (level), so we must process nodes grouped by their depth.

2. Step 2: Identify BFS level order traversal as the correct pattern

   BFS naturally processes nodes level by level, allowing us to track and update the maximum value for each level efficiently.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   BFS processes nodes level-wise, perfect for per-level max tracking [OK]

Hint: Level order traversal matches per-level processing [OK]

Common Mistakes:

• Using DFS without level tracking loses level grouping
• Greedy picks largest child but misses other nodes at same level
• DP without level info cannot isolate per-level max

3. Consider the following code snippet implementing the optimal approach to find the maximum width of a binary tree. Given the tree below: 1 / \ 3 2 / \ 5 9 What is the value of max_width returned by the function?

easy

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Trace first level

   Root node 1 at index 0; max_width updated to 1.

2. Step 2: Trace second level

   Nodes 3 and 2 at indices 0 and 1 (after normalization). Width = 1 - 0 + 1 = 2.

3. Step 3: Trace third level

   Nodes 5 and 9 at indices 0 and 3 (after normalization). Width = 3 - 0 + 1 = 4.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Maximum width is 4 at the last level [OK]

Hint: Normalize indices per level to compute width correctly [OK]

Common Mistakes:

• Not normalizing indices before computing width
• Counting nodes instead of width including null gaps
• Off-by-one error in width calculation

4. The following code attempts to perform vertical order traversal using BFS and column tracking. Identify the line containing the subtle bug that causes incorrect grouping of nodes by column.

from collections import deque
from sortedcontainers import SortedDict

def verticalOrder(root):
    if not root:
        return []
    col_table = SortedDict()
    queue = deque([(root, 0)])
    while queue:
        node, col = queue.popleft()
        if col in col_table:
            col_table[col] = []  # Bug here
        col_table[col].append(node.val)
        if node.left:
            queue.append((node.left, col - 1))
        if node.right:
            queue.append((node.right, col + 1))
    return list(col_table.values())

medium

A. Line: if not root: return []

B. Line: if col in col_table: col_table[col] = []

C. Line: queue.append((node.left, col - 1))

D. Line: queue = deque([(root, 0)])

Solution

1. Step 1: Analyze col_table initialization

   The condition resets col_table[col] to empty list if col already exists, erasing previous nodes in that column.

2. Step 2: Understand impact on output

   This causes nodes in the same column to be overwritten, breaking vertical grouping.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Correct code initializes list only if col not in col_table [OK]

Hint: Resetting list on existing key erases previous nodes [OK]

Common Mistakes:

• Confusing 'if col not in col_table' with 'if col in col_table'
• Not initializing column list properly

5. Suppose the vertical order traversal problem is extended so that nodes can appear multiple times if they have the same column and row but different parents (e.g., in a graph-like structure with shared nodes). Which modification to the BFS with column tracking approach correctly handles this scenario?

hard

A. Modify BFS to track (node, column, row) and store nodes in a list keyed by (column, row), then sort by parent ID

B. Use DFS with sorting by column, row, and parent to order nodes

C. Use BFS with column tracking but ignore row and parent, just append nodes as visited

D. Use BFS with column tracking and a visited set to avoid revisiting nodes

Solution

1. Step 1: Understand problem extension

   Nodes can appear multiple times if they share column and row but differ by parent, so row and parent info must be tracked.

2. Step 2: Evaluate modifications

   Tracking (node, column, row) and sorting by parent ID ensures correct ordering. Ignoring row or parent loses correctness. Visited set prevents duplicates but may skip valid repeats.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Tracking row and parent allows correct ordering of duplicates [OK]

Hint: Track row and parent to distinguish duplicates in BFS [OK]

Common Mistakes:

• Ignoring row or parent info leads to incorrect grouping
• Using visited set that skips valid repeated nodes