Leetcode: Maximum Depth of N-ary Tree

Posted on July 19, 2018September 10, 2019 by braindenny

Maximum Depth of N-ary Tree

Similar Problems:

CheatSheet: Leetcode For Code Interview
Tag: #treetraversal, #recursive, #dfs, #bfs

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.

For example, given a 3-ary tree:

We should return its max depth, which is 3.

Note:

The depth of the tree is at most 1000.
The total number of nodes is at most 5000.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: recursive

## Blog link: https://code.dennyzhang.com/maximum-depth-of-n-ary-tree
## Basic Ideas: recursive
##
## Complexity: Time O(n), Space O(n)

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, children):
        self.val = val
        self.children = children
"""
class Solution(object):
    def maxDepth(self, root):
        """
        :type root: Node
        :rtype: int
        """
        if root is None: return 0
        max_height = 0
        for child in root.children:
            max_height = max(max_height, self.maxDepth(child))
        return max_height+1

Solution: bfs

## Basic Ideas: bfs
##
## Complexity: Time O(n), Space O(n)

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, children):
        self.val = val
        self.children = children
"""
class Solution(object):
    def maxDepth(self, root):
        """
        :type root: Node
        :rtype: int
        """
        if root is None: return 0
        import collections
        queue = collections.deque()
        queue.append(root)
        level = 0
        while len(queue) != 0:
            level += 1
            for i in xrange(len(queue)):
                node = queue.popleft()
                for child in node.children:
                    queue.append(child)
        return level

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