
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Find Length of Longest Path with Even Sum in Python
Suppose we have a binary tree. we have to find the length of the longest path whose sum is an even number.
So, if the input is like image, then the output will be 5, as the path is like [5, 2, 4, 8, 5], sum = 24(even).
To solve this, we will follow these steps −
- Define a function dfs() . This will take node
- if node is null, then
- return a pair (0, -inf)
- (left_0, left_1) := dfs(left of node)
- (right_0, right_1) := dfs(right of node)
- if value of node is odd, then
- ans := maximum of ans, (left_1 + right_0 + 1) and (left_0 + right_1 + 1)
- return a pair (maximum of (left_1 + 1), (right_1 + 1) and 0) , maximum of (left_0 + 1) and (right_0 + 1))
- otherwise,
- ans := maximum of ans, (left_0 + right_0 + 1) and (left_1 + right_1 + 1)
- return a pair (maximum of (left_0 + 1), (right_0 + 1), 0) , maximum of (left_1 + 1), (right_1 + 1))
- From the main method do the following −
- ans := 0
- dfs(root)
- return ans
Example (Python)
Let us see the following implementation to get better understanding −
class TreeNode: def __init__(self, data, left = None, right = None): self.val = data self.left = left self.right = right class Solution: def solve(self, root): def dfs(node): if not node: return 0, float("-inf") left_0, left_1 = dfs(node.left) right_0, right_1 = dfs(node.right) if node.val & 1: self.ans = max(self.ans, left_1 + right_0 + 1, left_0 + right_1 + 1) return max(left_1 + 1, right_1 + 1, 0), max(left_0 + 1, right_0 + 1) else: self.ans = max(self.ans, left_0 + right_0 + 1, left_1 + right_1 + 1) return max(left_0 + 1, right_0 + 1, 0), max(left_1 + 1, right_1 + 1) self.ans = 0 dfs(root) return self.ans ob = Solution() root = TreeNode(2) root.left = TreeNode(5) root.right = TreeNode(4) root.right.left = TreeNode(8) root.right.right = TreeNode(2) root.right.left.left = TreeNode(5) print(ob.solve(root))
Input
root = TreeNode(2) root.left = TreeNode(5) root.right = TreeNode(4) root.right.left = TreeNode(8) root.right.right = TreeNode(2) root.right.left.left = TreeNode(5)
Output
5
Advertisements