Maximum Depth of N-ary Tree LeetCode Solution

Problem – Maximum Depth of N-ary Tree LeetCode Solution

Given a n-ary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See 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.

Maximum Depth of N-ary Tree LeetCode Solution in Java

public int maxDepth(Node root) {
    if(root == null) return 0;
    
    Queue<Node> queue = new LinkedList<>();
    queue.offer(root);
    
    int depth = 0;
    
    while(!queue.isEmpty())
    {
        int size = queue.size();
        
        for(int i = 0; i < size; i++)
        {
            Node current = queue.poll();
            for(Node child: current.children) queue.offer(child);
        }
        
        depth++;
    }
    
    return depth;
}

Maximum Depth of N-ary Tree LeetCode Solution in C++

class Solution {
public:
    int maxDepth(Node* root) {
        if (root == nullptr) return 0;
        int depth = 0;
        for (auto child : root->children) depth = max(depth, maxDepth(child));
        return 1 + depth;
    }
};

Maximum Depth of N-ary Tree LeetCode Solution in Python

"""
# Definition for a Node.
class Node:
    def __init__(self, val, children):
        self.val = val
        self.children = children
"""
class Solution:
    def maxDepth(self, root: 'Node') -> int:
        # Base case
        if root == None:
            return 0
        # Depth level of the tree
        depth = 0
        
        # Loops through children array
        for child in root.children:
            # Compares current depth of depth with a new level of depth 
            # Sets the biggest value to variable depth
            depth = max(depth, self.maxDepth(child))
        
        # As going deeper into the tree increases depth by 1
        print ('root ' + str(root.val) + ' depth ' + str(depth + 1))
        return depth + 1 
    
    # So for the first test case it'll look like
    # first call is the call with root 1
    # depth = 0 -> second call with child 3 -> 
    # it has 2 children so there'll be two calls from for loop
    # call with child 5 -> sets the for loop with child of 5 which is None (Null)
    # this call hits base case and returns 0
    # the same happens with child 6
    # call to child 6 from the for loop -> child of 6 is None -> returns 0
    # depth = max(0, 0) -> depth = 0
    # return to one level higher -> call to child 5 reaches return statement ->
    # depth = 1
    # call to child 6 reaches return statement and it returns 1 ->
    # current depth is 1 because child 5 set it -> max(1, 1) -> depth = 1
    # return to one more level higher
    # call to child 3 reaches return statement ->
    # depth was equal to 1 now it becomes 2 (depth + 1)
    # now it's turn to children 2 and 4 to be called
    # the story is exactly the same as with 5 and 6 only depth is now 2 ->
    # both 2 and 4 return depth 1 that's why it's rejected by max(2, 1)
    # depth remains 2
    # and finally everything returns to the very first call with root 1 
    # it reaches return statement and returns 2+1=3
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