Leetcode #222: Count Complete Tree Nodes
In this guide, we solve Leetcode #222 Count Complete Tree Nodes in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Problem Statement
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Bit Manipulation, Tree, Binary Search, Binary Tree
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: root = [1,2,3,4,5,6]
Output: 6
Python Solution
# Definition for a binary tree node.
# 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 root is None:
return 0
return 1 + self.countNodes(root.left) + self.countNodes(root.right)
Complexity
The time complexity is , and the space complexity is , where is the number of nodes in the tree. The space complexity is , where is the number of nodes in the tree.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.