543. Diameter of Binary Tree

[#binary-tree, #dfs]

## description

## 543. 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:

plain text

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:

plain text

Input: root = [1,2]
Output: 1

–
The number of nodes in the tree is in the range [1, 104].
–
-100 <= Node.val <= 100

Constraints:

## notes

### Intuition

The diameter passes through some node where the path goes down through its left and right subtrees. At each node, left_height + right_height is a candidate for the diameter. Track the maximum across all nodes.

### Implementation

Use DFS that returns the height of each subtree while updating a global result. Base case: null node returns 0. For each node, compute left and right heights recursively. Update the global maximum with left + right (the path through this node). Return 1 + max(left, right) as this subtree's height.

### Edge-cases

In Python, use nonlocal to modify variables from an enclosing scope within a nested function. The diameter could be entirely within one subtree.

–
Time: O(n) — visit each node once
–
Space: O(n) — recursion stack depth in worst case

### Complexity

## solution

python

from typing import Optional
from lcutils import TreeNode

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        res = 0
        def dfs(root: Optional[TreeNode]) -> int:
            nonlocal res
            if not root:
                return 0
            left = dfs(root.left)
            right = dfs(root.right)
            res = max(res, left + right)
            return 1 + max(left, right)

        dfs(root)
        return res
        



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()

--EOF--