main.py

# Source: https://leetcode.com/problems/count-complete-tree-nodes
# Title: Count Complete Tree Nodes
# Difficulty: Easy
# Author: Mu Yang <http://muyang.pro>

################################################################################################################################
# Given the `root` of a **complete** binary tree, return the number of the nodes in the tree.
#
# According to **Wikipedia**, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between `1` and `2^h` nodes inclusive at the last level `h`.
#
# Design an algorithm that runs in less than `O(n)` time complexity.
#
# **Example 1:**
# https://assets.leetcode.com/uploads/2021/01/14/complete.jpg
#
# ```
# Input: root = [1,2,3,4,5,6]
# Output: 6
# ```
#
# **Example 2:**
#
# ```
# Input: root = []
# Output: 0
# ```
#
# **Example 3:**
#
# ```
# Input: root = [1]
# Output: 1
# ```
#
# **Constraints:**
#
# - The number of nodes in the tree is in the range `[0, 5 * 10^4]`.
# - `0 <= Node.val <= 5 * 10^4`
# - The tree is guaranteed to be **complete**.
#
################################################################################################################################


from typing import Optional


class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


class Solution:
    def countNodes(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        return 1 + self.countNodes(root.left) + self.countNodes(root.right)