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0559 - Maximum Depth of N-ary Tree (Easy)

https://leetcode.com/problems/maximum-depth-of-n-ary-tree/

Problem Statement​

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, 10^4].
  • The depth of the n-ary tree is less than or equal to 1000.

Approach 1: DFS​

We should use the DFS template from DFS Guide. Instead of looking at the left and right child, we look at all children in an n-ary tree. We find the deepest child and set it to the current largestDepth, then we add one for counting the depth of the current level.

Note: A binary tree means each node at most has 2 children (binary means 2). An n-ary tree means each node at most has n children

Written by @heiheihang
def maxDepth(self, root: 'Node') -> int:
        
        def dfs(node):
            #there is a possibility the root is None
            if(node is None):
                return 0
                
            #initialize the largest depth
            largestDepth = 0
            
            #iterate each child, and update the largest depth if appropriate
            for child in node.children:
                largestDepth = max(largestDepth, dfs(child))
            
            #don't forget to add the current level height!    
            return largestDepth + 1
        
        #we start iterating from the root
        return dfs(root)