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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

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

- if flag is same as True, then
- 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

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