Adding Efficient Solution for Problem 543 - Diameter BT

Author: Vanditg

Diameter_BT/EfficientSolution.py

@@ -0,0 +1,31 @@
+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 543
+## Problem Name: Diameter of Binary Tree         
+##===================================
+#
+#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].
+class Solution:
+    def diameterOfBinaryTree(self, root):
+        self.tmp = 0	#Initialize tmp = 0
+        length = self.dfs(root)	#Initialize length, which is after running dfs, we'll get the count 
+        return tmp	#We'll return tmp 
+    def dfs(self, node):	#Define DFS
+        if node is None:	#Condition-check: If node is empty	 
+            return 0	#We return 0
+        l, r = self.dfs(node.left), self.dfs(node.right)	#Initialize l and r 
+        self.tmp = max(self.tmp, l + r)	#Update tmp by finding max between tmp and l + r
+        return 1 + max(l, r)	#Return 1 + max between l and r