543.py

# Name: Diameter of Binary Tree
# Link: https://leetcode.com/problems/diameter-of-binary-tree/
# Method: Solving tidbit
# Time: O(n)
# Space: O(d)
# Difficulty: Easy


# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution:

    cmax = 0

    def rec_dia(self, node):
        if not node:
            return 0

        depth_l = self.rec_dia(node.left)
        depth_r = self.rec_dia(node.right)

        dia_at = depth_l + depth_r
        print(f"At node {node.val} diameter is {dia_at}")
        if dia_at > self.cmax:
            self.cmax = dia_at

        return 1 + max(depth_l, depth_r)

    def diameterOfBinaryTree(self, root: TreeNode) -> int:
        self.cmax = 0
        self.rec_dia(root)
        return self.cmax


if __name__ == "__main__":
    root = TreeNode(1)
    root.left = TreeNode(2)
    root.left.left = TreeNode(3)

    root.right = TreeNode(4)

    root2 = TreeNode(0)
    root2.left = TreeNode(1)
    sol = Solution()
    # print(sol.diameterOfBinaryTree(root))
    print(sol.diameterOfBinaryTree(root2))