题解 | #二叉树的直径#

由于求的是树的边的最长长度,则递归过程由左 -> 右 -> 根 顺序进行,并在递归过程中记录 左子树直径 和 右子树直径的和的最大值

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @return int整型
#
class Solution:
    def __init__(self):
        self.path = 0
    
    def diameterOfBinaryTree(self , root: TreeNode) -> int:
        # write code here
        if not root:
            return 0
        self.subBinaryTree(root)
        return self.path
    
    def subBinaryTree(self, root: TreeNode) -> int:
        if not root:
            return 0
        l = self.subBinaryTree(root.left)
        r = self.subBinaryTree(root.right)
        self.path = max(self.path, l + r)
        return max(l, r) + 1