LeetCode 543 – Python 3 (Week 4 – 05)

Solution 1

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def diameterOfBinaryTree(self, root: TreeNode) -> int:
        
        self.ans = 0
        
        def depth(node):
            if not node: return 0
            L = depth(node.left)
            R = depth(node.right)
            self.ans = max(self.ans, L+R)
            return max(L, R) + 1
            
        depth(root)
        
        return self.ans

Time complexity is O(n). Space complexity is O(n).

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