0543.Diameter of Binary Tree

Link: 543. Diameter of Binary Tree
Track: NeetCode150

Question

Restate the problem

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Method 1
Method 2

Method

Approach

At each node:

1. left = dfs(node.left)
2. right = dfs(node.right)
3. update answer: diameter = max(diameter, left + right)
4. return height to parent: max(left, right) + 1

Complexity

• Time Complexity: O(n) (visit each node once)
• Space Complexity: O(h) recursion stack (worst O(n), balanced O(log n))

Edge Case

Code

# 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 diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        self.diameter = 0
 
        def dfs(node):
            if not node:
                return 0
 
            left = dfs(node.left)
            right = dfs(node.right)
            
            self.diameter = max(self.diameter, left + right)
            
            return max(left, right) + 1
        
        dfs(root)
        return self.diameter

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History

• Feb-22-2026 Solved, count length, not node
• Jan-19-2026 Peeked, no idea, use global variable