<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>

class Solution:
    def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        
        stack = []
        node_to_height = {}
        max_diameter = 0
        
        # Start DFS with the root node
        current = root
        last_visited = None
        
        while stack or current:
            if current:
                stack.append(current)
                current = current.left
            else:
                peek_node = stack[-1]
                # If the right child exists and hasn't been processed yet
                if peek_node.right and last_visited != peek_node.right:
                    current = peek_node.right
                else:
                    stack.pop()
                    left_height = node_to_height.get(peek_node.left, 0)
                    right_height = node_to_height.get(peek_node.right, 0)
                    
                    # Calculate the diameter passing through this node
                    max_diameter = max(max_diameter, left_height + right_height)
                    
                    # Store the height of this node
                    node_to_height[peek_node] = 1 + max(left_height, right_height)
                    
                    # Mark this node as visited
                    last_visited = peek_node
        
        return max_diameter           
18500