222. Count Complete Tree Nodes

Edited by Charles

Easy · Python · Binary Tree

Problem

Read the original problem on LeetCode.

The official page contains the problem statement, examples, and constraints.

Explanation, examples & other approaches

Core idea

Iteration:

Reference to 144. Binary Tree Preorder Traversal

Original study notes follow, including intermediate snippets and earlier attempts.

Iteration:

Reference to 144. Binary Tree Preorder Traversal

# 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
from collections import deque

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

        dq = deque([root])
        result = 0

        while dq:
            n = len(dq)

            for _ in range(n):
                node = dq.popleft()
                result += 1

                if node.left:
                    dq.append(node.left)
                if node.right:
                    dq.append(node.right)

        return result

Recursion

# 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:
        return self.traverse(root)

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

        leftSum = traverse(node.left)
        rightSum = traverse(node.right)

        return leftSum + rightSum + 1

Recursion with O(logN * logN)

Study record

Field	Value
Attempts	1
Mastery	❚
The Latest Attempt	January 13, 2026

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:
        left, right = root, root
        leftHeight, rightHeight = 0, 0

        while left:
            left = left.left
            leftHeight += 1

        while right:
            right = right.right
            rightHeight += 1

        if leftHeight == rightHeight:
            return pow(2, leftHeight) - 1

        return 1 + self.countNodes(root.left) + self.countNodes(root.right)

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