Maximum Depth of N-ary Tree

Problem Statement

Given an "n-ary" tree, where each node may have zero or more children, the challenge is to determine its maximum depth. The maximum depth is defined as the length of the longest path from the root node down to the farthest leaf node. Notably, this type of tree generalizes the concept of a binary tree to structures where each node can have multiple children.

The tree is presented in a serialized form based on level order traversal, with each group of children separated by a null value as shown in the examples. This structured approach allows for the representation of the tree's node-to-node connections and their hierarchical relationships.

Examples

Example 1

Input:

root = [1,null,3,2,4,null,5,6]

Output:

3

Example 2

Input:

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

Output:

5

Constraints

• The total number of nodes is in the range [0, 104].
• The depth of the n-ary tree is less than or equal to 1000.

Approach and Intuition

To solve the problem of finding the maximum depth in an n-ary tree, understanding the tree's traversal is crucial. Here is a breakdown based on the provided examples and constraints:

1. Tree Traversal and Representation:

   • The n-ary tree is represented in a serialized, level-order format. In this format, groups of children are separated by null values which makes visualizing the tree structure straightforward when parsing the array from left to right.

2. Understanding the Examples:

   • In the first example, the maximum depth is 3. This can be visualized by acknowledging the tree spreads across three levels, from the root node 1 down to the leaf nodes 5 and 6.
   • The second example has a maximum depth of 5, indicating a larger tree structure with more levels.

3. Building Intuition:

   • Identifying the maximum depth involves traversing each path from the root node to each leaf node and determining which path is the longest.
   • Given the constraints, with a maximum node count of 10,000 and depth limit of 1000, algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) are suitable. Using either method will efficiently cope with these size constraints.

4. Algorithm Selection:

   • DFS: This recursive approach explores as deep as possible along each branch before backtracking. It’s particularly effective for this problem because it naturally finds the deepest path by venturing to the bottom-most child before considering alternatives.
   • BFS: Alternatively, BFS uses a queue to explore the tree level by level, counting each level as it progresses which directly gives the depth of the tree.

To implement, one could either recursively determine the depth by examining all children and recursively calculating the depth for each subtree and taking the maximum, or iteratively using a queue to count levels. Both strategies effectively use the tree structure and the constraints to efficiently determine the maximum depth.

Solutions

class Solution {
  public int calculateMaxDepth(Node rootNode) {
    Queue<Pair<Node, Integer>> queueNodes = new LinkedList<>();
    if (rootNode != null) {
      queueNodes.add(new Pair(rootNode, 1));
    }

    int maxDepth = 0;
    while (!queueNodes.isEmpty()) {
      Pair<Node, Integer> currentItem = queueNodes.poll();
      rootNode = currentItem.getKey();
      int tempDepth = currentItem.getValue();
      if (rootNode != null) {
        maxDepth = Math.max(maxDepth, tempDepth);
        for (Node childNode : rootNode.children) {
          queueNodes.add(new Pair(childNode, tempDepth + 1));
        }
      }
    }
    return maxDepth;
  }
}