Program to find length of longest alternating path of a binary tree in python

Suppose we have a binary tree, we have to find the longest path that alternates between left and right child and going down.

So, if the input is like

then the output will be 5 as the alternating path is [2, 4, 5, 7, 8].

To solve this, we will follow these steps:

• if root is null, then
  • return 0
• Define a function dfs() . This will take node, count, flag
• if node is not null, then
  • if flag is same as True, then
    • a := dfs(left of node, count + 1, False)
    • b := dfs(right of node, 1, True)
  • otherwise when flag is same as False, then
    • a := dfs(right of node, count + 1, True)
    • b := dfs(left of node, 1, False)
  • return maximum of a, b
• return count
• From the main method do the following:
• a := dfs(left of root, 1, False)
• b := dfs(right of root, 1, True)
• return maximum of a, b

Let us see the following implementation to get better understanding:

Example

Live Demo

class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.data = data
      self.left = left
      self.right = right

class Solution:
   def solve(self, root):
      if not root:
         return 0

      def dfs(node, count, flag):
         if node:
            if flag == True:
               a = dfs(node.left, count + 1, False)
               b = dfs(node.right, 1, True)
            elif flag == False:
               a = dfs(node.right, count + 1, True)
               b = dfs(node.left, 1, False)

            return max(a, b)
      return count

      a = dfs(root.left, 1, False)
      b = dfs(root.right, 1, True)

   return max(a, b)

ob = Solution()
root = TreeNode(2)
root.left = TreeNode(3)
root.right = TreeNode(4)
root.right.left = TreeNode(5)
root.right.right = TreeNode(6)
root.right.left.right = TreeNode(7)
root.right.left.right.left = TreeNode(8)
print(ob.solve(root))

Input

root = TreeNode(2)
root.left = TreeNode(3)
root.right = TreeNode(4)
root.right.left = TreeNode(5)
root.right.right = TreeNode(6)
root.right.left.right = TreeNode(7)
root.right.left.right.left = TreeNode(8)

Output

5