python/problem_543/README.md

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+# 543. Diameter of Binary Tree
+
+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:
+
+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

python/problem_543/solution1.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:
+        diameter = 0
+
+        def longest_path(node):
+            # nonlocal to access the diameter when doing recursion
+            nonlocal diameter
+            # if no child
+            if not node:
+                return 0
+
+            # calculate the longest path for left and right
+            left_longest = longest_path(node.left)
+            right_longest = longest_path(node.right)
+
+            # check if its bigger than diameter
+            diameter = max(diameter, left_longest + right_longest)
+
+            # add the edge to parent while returning
+            return max(left_longest, right_longest) + 1
+
+        # do post-tree traversal
+        longest_path(root)
+        return diameter