Populating Next Right Pointers II (Any Tree)

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Problem

Imagine you have a complex organizational chart where each employee can have any number of direct reports, and you want to connect each employee to their immediate colleague on the right at the same level to facilitate quick lateral communication.

Given a binary tree where each node has an additional pointer called 'next', populate each node's next pointer to point to its next right node on the same level. If there is no next right node, the next pointer should be set to NULL. The tree can be any binary tree (not necessarily perfect or complete). You must solve the problem using constant extra space (excluding recursion stack).

The number of nodes in the tree is in the range [0, 10^5].Node values are arbitrary and do not affect the linking.You may only use constant extra space (excluding recursion stack).

Edge cases: Empty tree → output is NULLTree with only one node → next pointer remains NULLTree where some nodes have only left children → next pointers skip missing right siblings

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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 immediate siblings, missing nodes separated by null children.Greedy approach connecting only left child to right child without scanning next pointers at current level.✅ Use a dummy node and traverse current level using next pointers to connect all nodes at next level.

Wrong: Next pointer of last node at each level is not set to NULL.Off-by-one error forgetting to set prev.next = NULL after processing each level.✅ Explicitly assign prev.next = NULL at the end of each level traversal.

Wrong: Non-null output for empty tree input.Missing null check for root at start of function.✅ Add if root is None: return None at the beginning.

Wrong: Next pointers incorrectly assigned for single node tree.Assigning next pointers without checking if siblings exist.✅ Ensure next pointers are assigned only between existing sibling nodes.

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

✓

Test Cases

Try testing your solution on empty and single node trees to catch base case bugs.

Consider tricky cases where nodes have missing siblings or are not immediate neighbors.

Optimize your solution to avoid extra space and ensure it runs efficiently on large inputs.

t1_01basic

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

Expected[1,"#",2,3,"#",4,5,7,"#"]

⏱ Performance - must finish in 2000ms

At level 1, 1's next is NULL. At level 2, 2's next is 3, 3's next is NULL. At level 3, 4's next is 5, 5's next is 7, 7's next is NULL.

💡 Use level order traversal to connect nodes at each level.

💡 Track the previous node at the current level to set its next pointer.

💡 Set the last node's next pointer at each level to NULL explicitly.

Why it failed: Failed to connect next pointers correctly at all levels. Ensure you reset the previous pointer at each level and set the last node's next to NULL. This is the canonical example test.

✓ Correctly connected all next pointers as per the canonical example.

t1_02basic

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

Expected[1,"#",2,3,"#",4,5,"#"]

⏱ Performance - must finish in 2000ms

At level 1, 1's next is NULL. At level 2, 2's next is 3, 3's next is NULL. At level 3, 4's next is 5, 5's next is NULL. Nodes with missing siblings are skipped correctly.

💡 Handle missing children by skipping null nodes when connecting next pointers.

💡 Use a dummy node to build the next level's linked list efficiently.

💡 Traverse current level using next pointers instead of a queue for O(1) space.

Why it failed: Incorrectly connected next pointers when nodes have missing children. Make sure to skip null children and connect only existing nodes. This is a standard case with missing children.

✓ Successfully handled missing children and connected next pointers correctly.

t2_01edge

Input{"root":null}

Expectednull

⏱ Performance - must finish in 2000ms

Empty tree input should return null as there are no nodes to connect.

💡 Check if the root is null at the start.

💡 Return null immediately if the tree is empty.

💡 No connections needed when tree is empty.

Why it failed: Returned non-null for empty tree input. Fix by adding a null check at the start and returning null immediately. This is the empty tree edge case.

✓ Correctly handled empty tree by returning null.

t2_02edge

Input{"root":[1]}

Expected[1,"#"]

⏱ Performance - must finish in 2000ms

Single node tree's next pointer remains NULL as there are no siblings.

💡 For single node, next pointer should remain null.

💡 Ensure you do not assign next pointers when only one node exists.

💡 Check that your code handles single node without errors.

Why it failed: Incorrectly assigned next pointer for single node tree. Fix by ensuring next pointers are only assigned between siblings. This is the single node edge case.

✓ Correctly handled single node with next pointer as null.

t2_03edge

Input{"root":[1,2,null,3,null,4]}

Expected[1,"#",2,"#",3,"#",4,"#"]

⏱ Performance - must finish in 2000ms

Tree with only left children; next pointers skip missing right siblings and connect left children correctly.

💡 Traverse each level carefully to connect nodes even if some siblings are missing.

💡 Use a dummy node to chain nodes at the next level regardless of left or right child.

💡 Do not assume presence of right child; always check before connecting.

Why it failed: Failed to connect next pointers correctly when nodes have only left children. Fix by chaining all existing children regardless of side. This is the left-only children edge case.

✓ Correctly connected next pointers in left-only children tree.

t3_01corner

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

Expected[1,"#",2,3,"#",4,5,"#",6,"#"]

⏱ Performance - must finish in 2000ms

Greedy approach that connects only immediate siblings fails here; next pointers must skip nulls and connect nodes at same level even if not immediate siblings.

💡 Do not connect nodes greedily only if they are immediate siblings.

💡 Traverse the current level using next pointers to find the next available node.

💡 Use a dummy node to build the next level's linked list correctly.

Why it failed: Greedy approach failed by connecting only immediate siblings and missing nodes separated by nulls. Fix by scanning next pointers at current level to find next valid node. This is the greedy trap test.

✓ Correctly handled non-immediate siblings by connecting all nodes at the same level.

t3_02corner

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

Expected[1,"#",2,3,"#",4,5,"#",6,"#"]

⏱ Performance - must finish in 2000ms

Confusing 0/1 vs unbounded problem: each node appears once and must be connected once; do not connect nodes multiple times or skip nodes thinking of unbounded repetition.

💡 Remember each node appears once; do not reuse nodes or connect multiple times.

💡 Use level order traversal or next pointer traversal to connect nodes exactly once.

💡 Avoid loops or repeated connections by careful pointer updates.

Why it failed: Incorrectly connected nodes multiple times or skipped nodes due to misunderstanding of node uniqueness. Fix by connecting each node exactly once per level. This is the 0/1 vs unbounded confusion test.

✓ Correctly connected each node exactly once per level.

t3_03corner

Input{"root":[1,2,3,4,5,6,7,null,null,8]}

Expected[1,"#",2,3,"#",4,5,6,7,"#",8,"#"]

⏱ Performance - must finish in 2000ms

Off-by-one errors cause last node's next pointer to be non-null or missing. The last node at each level must have next pointer set to NULL.

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

💡 Reset the previous pointer at the start of each level traversal.

💡 Explicitly assign prev.next = NULL after processing each level.

Why it failed: Last node's next pointer incorrectly set to non-null or missing. Fix by explicitly setting prev.next = NULL at each level end. This is the off-by-one corner case.

✓ Correctly set last node's next pointer to NULL at each level.

t4_01performance

Input{"root":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,null,null,20,21,null,null,24,25,null,null,null,null,null,null,32,33,null,null,40,41,null,null,null,null,null,null,64,65,null,null,null,null,null,null,128,129]}

Expectednull

⏱ Performance - must finish in 2000ms

Large balanced binary tree with 127 nodes (7 levels). O(n) time expected to complete within 2 seconds.

💡 Use O(n) BFS or pointer traversal to connect nodes efficiently.

💡 Avoid extra space beyond constant by using next pointers to traverse levels.

💡 Do not use recursive solutions that may cause stack overflow on large trees.

Why it failed: Solution timed out due to using O(n) space or exponential time approach. Fix by implementing O(n) time and O(1) extra space solution using next pointers. This is the performance test for O(n) complexity.

✓ Solution runs efficiently in O(n) time and O(1) extra space.

Practice

1. You are given a binary tree and asked to compute the average value of nodes on each level. Which algorithmic approach guarantees an optimal solution that processes nodes level-by-level and efficiently computes these averages?

easy

A. Depth-first search (DFS) without tracking levels, summing all nodes and dividing by total count.

B. Breadth-first search (BFS) using a queue to traverse nodes level-by-level, summing values per level and dividing by the number of nodes at that level.

C. Greedy approach that picks the maximum node value at each level and averages those maximums.

D. Dynamic programming that stores partial sums of subtrees and combines them bottom-up.

Solution

1. Step 1: Understand the problem requires level-wise aggregation

   The problem asks for averages per level, so nodes must be grouped by their depth in the tree.

2. Step 2: Identify BFS as the natural approach for level order traversal

   BFS uses a queue to process nodes level-by-level, allowing summing and counting nodes at each level before moving on.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   BFS naturally groups nodes by level, enabling correct average calculation [OK]

Hint: Level order traversal = BFS with queue [OK]

Common Mistakes:

• Using DFS without level tracking loses level grouping
• Greedy max per level ignores all other nodes
• Bottom-up DP is unnecessary and complex here

2. Consider the following code snippet for counting good nodes in a binary tree. Given the tree with root value 3, left child 1 (with left child 3), and right child 4 (with left child 1 and right child 5), what is the final returned count?

easy

A. 4

B. 6

C. 5

D. 3

Solution

1. Step 1: Trace stack operations and count updates

   Start with stack=[(3,3)]. Pop (3,3), count=1. Push right (4,3), left (1,3). Pop (1,3), node.val=1 < 3 no count. Push left (3,3). Pop (3,3), node.val=3 >=3 count=2. Pop (4,3), node.val=4 >=3 count=3. Push right (5,4), left (1,4). Pop (1,4), node.val=1 <4 no count. Pop (5,4), node.val=5 >=4 count=4.

2. Step 2: Final count after traversal

   Count is 4 after visiting all nodes.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Count matches expected good nodes in example [OK]

Hint: Track max_val updates carefully during stack pops [OK]

Common Mistakes:

• Off-by-one counting
• Ignoring max_val update before pushing children
• Mixing left and right child order

3. What is the time complexity of the optimal DFS approach for bottom-up level order traversal of a binary tree with n nodes?

medium

A. O(n log n) due to reversing the levels list at the end

B. O(n) for traversal plus O(h) for recursion stack space, where h is tree height

C. O(n) because each node is visited exactly once and reversal is O(n)

D. O(n^2) because appending to levels at each depth can be costly

Solution

1. Step 1: Analyze DFS traversal cost

   Each node is visited once, so traversal is O(n).

2. Step 2: Analyze reversal cost

   Reversing the levels list is O(L), where L ≤ n (number of levels), so O(n) worst case.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Traversal plus reversal both linear in n [OK]

Hint: DFS visits each node once; reversal is linear [OK]

Common Mistakes:

• Confusing reversal cost as O(n log n)
• Assuming quadratic due to nested lists

4. The following code attempts the optimal DFS approach for bottom-up level order traversal. Identify the line containing the subtle bug that causes incorrect output for an empty tree input.

def levelOrderBottom(root):
    levels = []
    def dfs(node, level):
        if not node:
            return
        if len(levels) == level:
            levels.append([])
        levels[level].append(node.val)
        dfs(node.left, level + 1)
        dfs(node.right, level + 1)
    dfs(root, 0)
    return levels[::-1]

medium

A. Line 10: dfs(root, 0)

B. Line 11: return levels[::-1]

C. Line 2: levels = []

D. Line 3: def dfs(node, level):

Solution

1. Step 1: Consider empty tree input (root = None)

   If root is None, dfs returns immediately without adding levels.

2. Step 2: Check what happens after dfs call

   levels remains empty, so return levels[::-1] returns [], which is correct and expected.

3. Step 3: Identify bug

   Returning levels[::-1] when levels is empty is correct; the subtle bug is missing an explicit check for root being None before calling dfs to avoid unnecessary calls.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Returning reversed empty list is correct; no bug in dfs call line [OK]

Hint: Returning reversed empty list is safe for empty tree [OK]

Common Mistakes:

• Assuming dfs call on None causes error
• Not handling empty tree explicitly but no failure occurs

5. Suppose the problem is extended so that the tree nodes can have multiple children (not just binary), and you need to check if the tree is symmetric around its center. Which modification to the BFS approach is necessary to correctly handle this case?

hard

A. Enqueue children in normal order and compare corresponding children pairs directly

B. Enqueue children in mirrored order and compare pairs from opposite ends of the children list

C. Use a recursive approach only, as BFS cannot handle multiple children

D. Check symmetry only at the root level and ignore deeper levels

Solution

1. Step 1: Understand symmetry in multi-child trees

   Symmetry means the first child on the left subtree mirrors the last child on the right subtree, second mirrors second last, etc.

2. Step 2: Modify BFS to enqueue children in mirrored order

   Enqueue children pairs from opposite ends of the children list to compare mirrored positions correctly.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Mirrored enqueue order for multiple children preserves symmetry checks [OK]

Hint: Multi-child symmetry requires mirrored child pair comparisons [OK]

Common Mistakes:

• Enqueuing children in normal order causing false positives
• Assuming BFS cannot be adapted for multi-child trees
• Ignoring deeper levels of the tree