## [543\. Diameter of Binary Tree](https://leetcode.com/problems/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.

**Example 1:**


**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`

算 Binary Tree 的直徑，dfs 解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    int diameter = 0;    
public:
    int diameterOfBinaryTree(TreeNode* root) {
        dfs(root);
        return diameter;
    }

    int dfs(TreeNode* root) {
        if(!root) return 0;
        // 走訪左邊的 tree
        int left = dfs(root->left);
        
        // 走訪右邊的 tree
        int right = dfs(root->right);
        
        // 算最大的直徑
        diameter = max(diameter, left + right);
        
        // 每次遞迴，要記得直徑 + 1
        return 1 + max(left, right);
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::