diameter-of-bt py3

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]
+