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# [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
}
```