Diameter of a Tree

#127 · Trees · Advanced · 120 XP · about 20 min

The diameter of a binary tree is the number of edges on the longest path between any two nodes, where a path follows parent-child links and never visits a node twice. The path does not have to pass through the root. Write diameter(root) that returns it. A tree with no nodes or a single node has diameter 0.

Each node is a Node with a value val and two children, left and right, which are None when missing. The provided build makes a tree from a list of values in level order, where None marks a missing child.

Examples

diameter(build([6, 3, 9, 1, 4]))  # returns 3
diameter(build([5, 8]))  # returns 1
diameter(build([7]))  # returns 0

The drawing shows the first tree. Its longest paths, 1, 3, 6, 9 and 4, 3, 6, 9, each have 3 edges.

Constraints

0 to 2,000 nodes.

Note

The longest path can stay entirely inside one subtree, away from the root.

# Provided: do not change

from collections import deque

class Node:

def __init__(self, val, left=None, right=None):

self.val = val

self.left = left

self.right = right

def build(values):

"""Build a tree from a level-order list; None marks a missing child."""

if not values or values[0] is None:

return None

root = Node(values[0])

queue = deque([root])

i = 1

while queue and i < len(values):

node = queue.popleft()

if values[i] is not None:

node.left = Node(values[i])

queue.append(node.left)

i += 1

if i < len(values) and values[i] is not None:

node.right = Node(values[i])

queue.append(node.right)

i += 1

return root

def diameter(root):

# Your code here

pass

print(diameter(build([6, 3, 9, 1, 4])))

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