April Day 11 done

problems/543/main.py

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+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+class Solution:
+    def diameterOfBinaryTree(self, root: TreeNode) -> int:
+        if root is not None:
+            l = self.height(root.left)
+            r = self.height(root.right)
+            ldiameter = self.diameterOfBinaryTree(root.left)
+            rdiameter = self.diameterOfBinaryTree(root.right)
+            return max(l + r, max(ldiameter, rdiameter))
+        return 0
+    
+    def height(self, tree):
+        if tree is None:
+            return 0
+        return 1 + max(self.height(tree.left), self.height(tree.right))
+
+s = Solution()
+print("Expected: 3")
+bt = TreeNode(1)
+bt.left = TreeNode(2)
+bt.right = TreeNode(3)
+bt.left.left = TreeNode(4)
+bt.left.right = TreeNode(5)
+print("Got:", s.diameterOfBinaryTree(bt))

problems/543/problem.txt

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+ Given a binary tree, you need to compute 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.
+
+Example:
+Given a binary tree
+
+          1
+         / \
+        2   3
+       / \     
+      4   5    
+
+Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
+
+Note: The length of path between two nodes is represented by the number of edges between them. 

problems/543/run.sh

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+#!/bin/bash
+
+python3 main.py