<p>Given the <code>root</code> of a binary tree, return <em>the length of the <strong>diameter</strong> of the tree</em>.</p>

<p>The <strong>diameter</strong> of a binary tree is the <strong>length</strong> of the longest path between any two nodes in a tree. This path may or may not pass through the <code>root</code>.</p>

<p>The <strong>length</strong> of a path between two nodes is represented by the number of edges between them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg" style="width: 292px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 3 is the length of the path [4,2,1,3] or [5,2,1,3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1,2]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

# 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:
        # 1st bsf
        waiting_list = [root]
        node = None
        parents = {}

        while waiting_list:
            node = waiting_list.pop(0)

            if node.left:
                parents[node.left] = node
                waiting_list.append(node.left)
            if node.right:
                parents[node.right] = node
                waiting_list.append(node.right)

        waiting_list = [node, ]
        level = -1
        visited = []

        while waiting_list:
            level += 1
            for _ in range(len(waiting_list)):
                node = waiting_list.pop(0)
                visited.append(node)

                if node.left and node.left not in visited:
                    waiting_list.append(node.left)
                if node.right and node.right not in visited:
                    waiting_list.append(node.right)
                if node in parents and parents[node] not in visited:
                    waiting_list.append(parents[node])
        
        return level
18200