二叉树的直径【二叉树】

Problem: 543. 二叉树的直径

思路 & 解题方法

二叉树递归

复杂度

时间复杂度:

$O(n)$

空间复杂度:

$O(1)$

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:
        if not root:
            return 0

        L = self.diameterOfBinaryTree(root.left)
        R = self.diameterOfBinaryTree(root.right)
        ans = 0
        if root.left and root.right:
            root.val = max(root.left.val, root.right.val) + 1
            ans = root.left.val + root.right.val
        elif not root.left and not root.right:
            root.val = 1
            ans = 0
        else:
            root.val = root.left.val + 1 if root.left else root.right.val + 1
            ans = root.val - 1

        return max([L, R, ans])