---
title: "543. Diameter of Binary Tree"
url: "https://laigary.com/interview/coding/543-diameter-of-binary-tree"
type: "note"
section: "coding"
date: "2023-01-29"
updated: "2025-03-24"
tags: ["Tree", "Postorder Traversal"]
---

# 543. Diameter of Binary Tree

[543\. Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/)

這一題和 [124\. Binary Tree Maximum Path Sum](/interview/coding/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。

```python
# 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。

```python
# 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

```