Problem Description

In this LeetCode problem, we are given a binary tree where each node has an additional pointer called next. This tree structure is different from the standard binary tree structure, which typically only includes left and right pointers to represent the node's children. The next pointer in each node should be used to point to its immediate right neighbor on the same level. If no such neighbor exists, the next pointer should be set to NULL.

To clarify, if we imagine the nodes being aligned at each level from left to right, the next pointer of a node should reference the adjacent node to the right within the same level. For the rightmost nodes of each level, since there is no node to their right, their next pointer will remain NULL as initialized.

Initially, all the next pointers of the nodes in the tree are set to NULL. Our task is to modify the tree such that all next pointers are correctly assigned according to the problem's rules.

Flowchart Walkthrough

Let's dive into the algorithm for the problem using the Flowchart. Here's how we proceed step by step:

Is it a graph?

• Yes: Although not explicitly a graph problem, the tree structure can be considered a graph where nodes are connected to their children and potentially their siblings.

Is it a tree?

• Yes: The problem involves a binary tree, even though it's not a perfect binary tree.

DFS

• Depth-First Search (DFS) is the suggested pattern when we are dealing with a tree structure to traverse or manipulate individual nodes deeply before moving to other branches.

Conclusion: According to the flowchart, using DFS is appropriate for a tree structure like the one in this problem. In this specific problem, DFS can help track and connect the next right pointers thoroughly from deepest nodes upward or vice versa, ensuring each node points to its right neighbor or null if it is the rightmost node in its level.

Intuition

The solution involves a level-order traversal (BFS-like approach) of the binary tree, where we iterate through nodes level by level. Since we need to connect nodes on the same level from left to right, we keep track of the first node on the new level that we visit while still iterating on the current level. This way, we have a reference when we need to move down a level.

To solve this, we need variables to keep track of:

1. The previous node on the current level (prev), so we can set its next pointer to the current node.
2. The first node on the next level (next), which is where we will move after finishing the current level.
3. The current node that we're attaching to the next pointer of the previous node (curr).

A helper function modify takes care of the linking process, linking the current node to the previous one if it exists, and also updating the next variable to point to the first node at the next lower level when moving down the tree.

We use a while loop to navigate through the levels, and inside the loop, we traverse the current level using the next pointers (which are already pointing to the right or are NULL for the rightmost nodes). As we go, we use the modify helper function to link the children of the current node properly. We then transition down to the next level of the tree using the next variable, which we've already set to the first node on that level.

The traversal repeats until there are no more levels to process (i.e., when we've processed the rightmost node at the lowest level of the tree), ensuring that all next pointers are appropriately linked.

Learn more about Tree, Depth-First Search, Breadth-First Search, Linked List and Binary Tree patterns.

Solution Approach

The implementation for this problem involves a clever usage of pointers to traverse the binary tree level by level without using additional data structures like queues, which are commonly used for level-order traversal. Here's a step-by-step walkthrough of the process using the given Python code:

1. Initialize Pointers: We start by creating a node pointer that points to the root of the tree. The loop that follows will process one level at a time. Two additional pointers, prev and next, are used to track the previous node and the first node on the next level, respectively.

2. Outer While Loop: The outer loop continues until node is None. This loop ensures that we visit all the levels of the binary tree.

3. Set Level Pointers: At the beginning of each iteration, we set prev and next to None. Here, prev will hold the last node we visited on the current level, and next will be updated to point to the first node of the next level as soon as we encounter it.

4. Inner While Loop: The inner loop processes each level. It continues while there is a node to process on the current level.

5. Helper Function - modify: Each iteration within the inner while loop calls the modify function with the node.left and node.right children. This function does three things:

   • Checks if the current child curr is None. If it is, we don't have anything to modify or link, so we return immediately.
   • The first non-None child encountered will be the starting node for the next level, and we update next to point to this child.
   • If there's a prev node, we link its next pointer to the current child, curr. This connects the previous child on the same level to the current one.
   • Assigns curr to prev because for the next child, curr will be the previous node.

6. Traverse Current Level: After the children of the current node are processed, we move to the next node on the same level by updating node = node.next.

7. Move to Next Level: Once the inner loop is done, we have processed all nodes on the current level, and next points to the first node of the next level. We set node = next to move down the tree and begin processing the next level.

8. Return Modified Tree: The process continues until all nodes have been processed. The outer loop ends, and the original root, which now has all its next pointers correctly set, is returned.

Throughout the process, we are effectively doing a BFS traversal without an auxiliary queue, using next pointers to navigate through the tree instead. The solution leverages the fact that we can link the next level's nodes while we are still on the current level, using the next pointers that we're setting along the way.

Example Walkthrough

To illustrate the solution approach, consider a binary tree with the following structure:

        1
       / \
      2   3
     / \   \
    4   5   7

We start by initiating a variable node to point to the root of the tree, in this case, the node with value 1. The variables prev and next are initialized to None. The outer while loop starts, indicating we're beginning with the top level of the tree.

• At the top level:

  • The inner while loop processes the top level.
  • The modify helper function is called first with node.left (2) and then with node.right (3). Since prev is None, we just update next to node 2 (the first child of this level).
  • prev is then updated to 2.
  • We connect 2 to 3 by setting 2.next to 3 through the modify function with node.right.
  • Since node 3 is the rightmost node at this level, its next pointer is left as None.
  • Finally, the inner loop concludes, as there are no more nodes on the top level.

• We switch levels by setting node to next, which is node 2, and prev and next to None.

• At the second level:

  • We start with node 2. Again, the inner loop processes the level.
  • The modify function is called first with node.left (4) and then with node.right (5). next is updated to 4, and prev to 4.
  • prev (4) is connected to 5 via 5's next pointer because 5 is node.right of 2.
  • On to node 3, its children are None and 7. So, modify skips the None child, and connects 5 (the prev) to 7, since 7 is node.right of 3.
  • The rightmost nodes (5 and 7) have their next pointers already as None, and they stay that way since there are no more nodes to their right.
  • After finishing node 3, the inner loop concludes, as there are no more nodes at this level with a non-NULL next pointer.

• We again switch levels by setting node to next, which is node 4. However, nodes 4, 5, and 7 have no children, so subsequent iterations of the outer loop will not modify any next pointers.

After the outer while loop exits, we have successfully connected all the next pointers:

        1 -> NULL
       / \
      2 -> 3 -> NULL
     / \    \
    4 ->5 -> 7 -> NULL

The tree's next pointers have been set up to link nodes together across each level from left to right, with the rightmost nodes pointing to NULL, as required by the problem. All the intended connections have been made while respecting the binary tree structure and the rule stating that only immediate right neighbors should be linked. The solution also ensures that no additional data structures are used by leveraging the next pointers and processing the tree level by level.

Solution Implementation

Time and Space Complexity

The given code is designed to connect each node to its next right node in a binary tree, making use of level order traversal.

Time Complexity:

The time complexity of the code is O(N) where N is the number of nodes in the binary tree. This is because the algorithm visits each node exactly once. During each visit, it only performs constant time operations such as setting the next pointer.

Each level of the tree is examined using a while loop which iterates through the nodes that are horizontally connected via the next pointer. For each node, it calls the modify function twice—once for the left child and once for the right child. However, these calls do not result in recursive function calls that would amplify the runtime since they only alter pointers if the children are not null.

Space Complexity:

The space complexity of the code is O(1) assuming that function call stack space isn't considered for the space complexity (as is common for such analysis). This is because the algorithm uses only a constant amount of extra space: a few pointers (curr, prev, next) to keep track of the current node and to link the next level nodes. It does not allocate any additional data structures that grow with the size of the input tree.

However, if we do consider recursive call stack then the space complexity would be O(H) where H is the height of the tree. This accounts for the call stack during the execution of the function when called recursively for each level of the tree. For a balanced binary tree, this would be O(log N), and for a completely unbalanced tree, it could be O(N) in the worst case. The provided code does not seem to use recursion, so the space complexity remains O(1).

Learn more about how to find time and space complexity quickly using problem constraints.