543. Diameter of Binary Tree
這一題和 124. Binary Tree Maximum Path Sum 很類似，都是一題後序遍歷的題目，其中遍歷的返回值並不是我們要的答案，答案是是要在後序遍歷的過程中，利用後序遍歷的求解的過程，找出我們要的答案。
和 124 一樣，我們可以先開啟觀察者模式，透過觀察者找出所有的圓周。
# 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 __init__(self):
self.diameter = 0
def diameterOfBinaryTree(self, root: TreeNode) -> int:
diameters= []
def helper(root):
if not root:
return 0
left = helper(root.left)
right = helper(root.right)
diameters.append(left + right)
return max(left, right) + 1
helper(root)
return max(diameters)
簡化成一個變數。
# 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 __init__(self):
self.diameter = 0
def diameterOfBinaryTree(self, root: TreeNode) -> int:
def helper(root):
if not root:
return 0
left = helper(root.left)
right = helper(root.right)
self.diameter = max(self.diameter, left + right)
return max(left, right) + 1
helper(root)
return self.diameter