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


## 题目

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.

**Example 1:**

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

```
Input: root = [1,2,3,4,5]
Output: 3
Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].

```

**Example 2:**

```
Input: root = [1,2]
Output: 1

```

**Constraints:**

- The number of nodes in the tree is in the range `[1, 104]`.
- `100 <= Node.val <= 100`

## 题目大意

给定一棵二叉树，你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点。

## 解题思路

- 简单题。遍历每个节点的左子树和右子树，累加从左子树到右子树的最大长度。遍历每个节点时，动态更新这个最大长度即可。

## 代码

```go
package leetcode

import (
	"github.com/halfrost/LeetCode-Go/structures"
)

// TreeNode define
type TreeNode = structures.TreeNode

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

func diameterOfBinaryTree(root *TreeNode) int {
	result := 0
	checkDiameter(root, &result)
	return result
}

func checkDiameter(root *TreeNode, result *int) int {
	if root == nil {
		return 0
	}
	left := checkDiameter(root.Left, result)
	right := checkDiameter(root.Right, result)
	*result = max(*result, left+right)
	return max(left, right) + 1
}

func max(a int, b int) int {
	if a > b {
		return a
	}
	return b
}
```