LeetCode 솔루션					분류
				
						[11/15] 222. Count Complete Tree Nodes
본문
222. Count Complete Tree Nodes
Medium
6649375Add to ListShareGiven 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:

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 * 104].
- 0 <= Node.val <= 5 * 104
- The tree is guaranteed to be complete.
Accepted
502,754
Submissions
846,550
관련자료
- 
			링크
			댓글 1
					
			학부유학생님의 댓글
- 익명
- 작성일
					
										
					Runtime: 193 ms, faster than 33.94% of Python3 online submissions for Count Complete Tree Nodes.
Memory Usage: 21.5 MB, less than 46.74% of Python3 online submissions for Count Complete Tree Nodes.
				
													
								Memory Usage: 21.5 MB, less than 46.74% of Python3 online submissions for Count Complete Tree Nodes.
# 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 not root: return 0
        
        l_depth = self.count_depth(root.left)
        r_depth = self.count_depth(root.right)
    
        if l_depth == r_depth:
            return pow(2, l_depth) + self.countNodes(root.right)
        else:
            return pow(2, r_depth) + self.countNodes(root.left)
    
    
    def count_depth(self, node):
        if not node: return 0
        return 1 + self.count_depth(node.left) 
								 
							








