Leetcode 429: N-ary Tree Level Order Traversal

Input: The input is serialized as a list where each node’s value is followed by its children, and groups of children are separated by null values.

Example: root = [10, null, 20, 30, 40, null, 50, 60]

Constraints:

• The total number of nodes is between [0, 104].

• The height of the tree is less than or equal to 1000.

Output: The output should be a list of lists where each sublist represents the nodes at a particular level.

Example: [[10], [20, 30, 40], [50, 60]]

Constraints:

• The output should group nodes by their levels.

Goal: The goal is to perform a level order traversal and group the nodes at each level into separate lists.

Steps:

• 1. Initialize an empty result list to hold the groups of nodes.

• 2. Traverse the tree level by level, using a queue to store nodes at each level.

• 3. For each node, add its value to the corresponding level’s group in the result list.

• 4. Continue the traversal until all nodes are processed.

Goal: The solution should handle a tree with a height of up to 1000 and up to 10^4 nodes efficiently.

Steps:

• The tree height is at most 1000.

• The number of nodes can be up to 10,000.

Assumptions:

• The tree structure is a valid n-ary tree.

• Each node can have any number of children.

• Input: Input: root = [1, null, 2, 3, 4, 5, null, null, 6, 7, null, 8, 9, 10, null, null, 11, null, 12, null, 13, null, null, 14]

• Explanation: The tree structure is divided into levels. At each level, we group nodes together, starting with the root node at level 0. The children of each node are grouped at the next level.

Approach: We will perform a level order traversal using a queue. The key is to handle the nodes at each level and collect their values separately in the result list.

Observations:

• A level order traversal ensures that we process each level of nodes one by one.

• Using a queue will allow us to process nodes level by level, ensuring we capture the correct ordering.

Steps:

• 1. Create an empty queue to store nodes to be processed.

• 2. Process the nodes in the queue and group their values by level.

• 3. For each node, add its children to the queue, ensuring the order is maintained.

Empty Inputs:

• An empty tree should return an empty list.

Large Inputs:

• Large trees with up to 10,000 nodes should be handled efficiently.

Special Values:

• Ensure correct behavior with nodes having no children (leaf nodes).

Constraints:

• The solution should handle edge cases such as an empty tree or a tree with only one node.

vector<vector<int>> levelOrder(Node* root) {
    if(!root) return {};
    vector<vector<int>> ans;
    levelOrderT(root, 0, ans);
    return ans;
}

void levelOrderT(Node* node, int level, vector<vector<int>> &ans) {
    if(level == size(ans)) {
        ans.push_back({node->val});
    } else {
        ans[level].push_back(node->val);
    }

    for(Node* child: node->children) {
        levelOrderT(child, level+1, ans);
    }
}

1 : Function Declaration

vector<vector<int>> levelOrder(Node* root) {

Declares the main function for level-order traversal, returning a 2D vector containing nodes grouped by level.

2 : Base Condition

    if(!root) return {};

Handles the edge case where the tree is empty, returning an empty result.

3 : Variable Initialization

    vector<vector<int>> ans;

Initializes a 2D vector to store the traversal result grouped by levels.

4 : Recursive Function Call

    levelOrderT(root, 0, ans);

Starts the recursive level-order traversal from the root node.

5 : Return Result

    return ans;

Returns the final 2D vector containing the level-order traversal result.

6 : Recursive Function Declaration

void levelOrderT(Node* node, int level, vector<vector<int>> &ans) {

Declares the recursive helper function for level-order traversal.

7 : Conditional Check

    if(level == size(ans)) {

Checks if the current level exists in the 2D vector; if not, creates a new level.

8 : Push

        ans.push_back({node->val});

Adds the current node's value to a new level in the result vector.

9 : Else Condition

    } else {

Handles the case where the level already exists in the result vector.

10 : Push

        ans[level].push_back(node->val);

Appends the current node's value to the existing level in the result vector.

11 : Condition End

Ends the conditional block for checking the level.

12 : Loop

    for(Node* child: node->children) {

Iterates through all children of the current node to process them recursively.

13 : Recursive Call

        levelOrderT(child, level+1, ans);

Calls the helper function for each child, incrementing the level.

Best Case: O(n), where n is the total number of nodes in the tree.

Average Case: O(n), since we process each node once.

Worst Case: O(n), since we must visit all nodes.

Description: The time complexity is linear with respect to the total number of nodes.

Best Case: O(n), since we store all nodes in the queue for processing.

Worst Case: O(n), where n is the total number of nodes.

Description: Space complexity is linear, as we store nodes at each level during traversal.

Leetcode 429: N-ary Tree Level Order Traversal

Solution to LeetCode 429: N-ary Tree Level Order Traversal Problem

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