---
description: Recursive Tree
---

# 543. Diameter of Binary Tree

Given the `root` of a binary tree, return _the length of the **diameter** of the tree_.

The **diameter** of a binary tree is the **length** of the longest path between any two nodes in a tree. This path may or may not pass through the `root`.

The **length** of a path between two nodes is represented by the number of edges between them.

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg)

<pre><code><strong>Input: root = [1,2,3,4,5]
</strong><strong>Output: 3
</strong><strong>Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: root = [1,2]
</strong><strong>Output: 1
</strong></code></pre>

&#x20;

**Constraints:**

* The number of nodes in the tree is in the range `[1, 10`<sup>`4`</sup>`]`.
* `-100 <= Node.val <= 100`

分析

求最长的线段，# of nodes - 1

递归函数不断更新最长边长，返回高度

```
from typing import Optional

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:
        res = 0
        def helper(root):
            nonlocal res
            if not root:
                return 0
            left = helper(root.left)
            right = helper(root.right)
            res = max(res, left + right) #count edges, not nodes
            return 1 + max(left, right) #return height(# of nodes)
        helper(root) 
        return res

```
