#+ANKI_DECK: study_deck_02
* TODO 0543. Diameter of Binary Tree :easy:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0543. Diameter of Binary Tree][0543. Diameter of Binary Tree]]
:END:

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:*


#+begin_src
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].
#+end_src


*Example 2:*


#+begin_src
Input: root = [1,2]
Output: 1
#+end_src


*Constraints:*

 - The number of nodes in the tree is in the range ~[1, 10^{4}]~.

 - ~-100 <= Node.val <= 100~

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python :lc-problem 543 :lc-lang python3
# 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:
#+end_src

** TODO C++
#+begin_src cpp :lc-problem 543
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int diameterOfBinaryTree(TreeNode* root) {
        
    }
};
#+end_src