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.

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);
    }
};

• 時間複雜度：
  $O(N)$
• 空間複雜度：
  $O(N)$