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+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:**
+
+![](https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg)
+
+    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, 104]`.
+*   `-100 <= Node.val <= 100`
+
+https://leetcode.com/problems/diameter-of-binary-tree/
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diff --git a/0543_diameter-of-binary-tree/python3/solution.py b/0543_diameter-of-binary-tree/python3/solution.py
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+# 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:
+        result = [0]
+        
+        def dfs(node):
+            # Leaf node height will be 1
+            # Empty node height will be -1
+            
+            if node is None: return -1
+            
+            lefth = dfs(node.left)
+            righth = dfs(node.right)
+            
+            # Diameter calc:
+            #
+            # Cause for current node will be pointing to both
+            # left and right nodes and we need to consider
+            # them (+2)
+            #
+            # e.g. consider this is leaf node, lefth and righth = -1
+            # so, current node's diameter should be 0 = 2 + -1 + -1
+            result[0] = max(result[0], 2 + lefth + righth)
+            
+            # return max height including the node itself
+            return 1 + max(lefth, righth)
+        
+        dfs(root)
+        
+        return result[0]
+            
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