# 222 Count Complete Tree Nodes

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 `2h` nodes inclusive at the last level `h`.

Design an algorithm that runs in less than `O(n)` time complexity.

Example 1:

``````     1
/   \
2     3
/ \   /
4   5 6

Input: root = [1,2,3,4,5,6]
Output: 6
``````

Example 2:

``````Input: root = []
Output: 0
``````

Example 3:

``````Input: root = 
Output: 1
``````

Constraints:

• The number of nodes in the tree is in the range `[0, 5 * 104]`.
• `0 <= Node.val <= 5 * 104`
• The tree is guaranteed to be complete.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 `````` ``````# 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: # count the depth of left-most and right-most nodes l, r = root, root lh, rh = 0, 0 while l: l = l.left lh += 1 while r: r = r.right rh += 1 # if left-most and right-most have the same depth # the current tree is a full binary tree if lh == rh: return 2 ** lh - 1 return 1 + self.countNodes(root.left) + self.countNodes(root.right)``````