Diameter of Binary Tree
Problem
https://leetcode.com/problems/diameter-of-binary-tree/
Given the root of a binary tree, return the length of the
diameter of the tree.
The diameter of a binary tree is the length of the longest path
between any two nodes in a tree. This path may or may not pass through
the root.
The length of a path between two nodes is represented by the number of edges between them.
Example 1:

Input: root = [1,2,3,4,5]
Output: 3
Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
Example 2:
Input: root = [1,2]
Output: 1
Constraints:
The number of nodes in the tree is in the range
[1, 10:sup:`4`].-100 <= Node.val <= 100
Pattern
Tree, Depth-First Search, Binary Tree
Approaches
Code
from __future__ import annotations
class TreeNode:
"""Node in a binary tree."""
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
@classmethod
def from_list(cls, vals: list[int | None]) -> TreeNode | None:
if not vals:
return None
root = TreeNode(vals[0])
queue = [root]
i = 1
while i < len(vals):
node = queue.pop(0)
if i < len(vals) and vals[i] is not None:
node.left = TreeNode(vals[i])
queue.append(node.left)
i += 1
if i < len(vals) and vals[i] is not None:
node.right = TreeNode(vals[i])
queue.append(node.right)
i += 1
return root
def diameterOfBinaryTree(root: TreeNode | None) -> int:
"""Return the diameter of the binary tree."""
result = 0
def dfs(node: TreeNode | None) -> int:
nonlocal result
if node is None:
return 0
left = dfs(node.left)
right = dfs(node.right)
result = max(result, left + right)
return 1 + max(left, right)
dfs(root)
return result
Test
>>> from diameter_of_binary_tree__dfs import TreeNode, diameterOfBinaryTree
>>> diameterOfBinaryTree(TreeNode.from_list([1, 2, 3, 4, 5]))
3
>>> diameterOfBinaryTree(TreeNode.from_list([1, 2]))
1
- class diameter_of_binary_tree__dfs.TreeNode(val=0, left=None, right=None)
Bases:
objectNode in a binary tree.