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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

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

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

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