DFS解法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private: 
    int max_len = 0;
public:
    int dfs(TreeNode* root) {
        // 叶节点返回深度0
        int right = (root->left != nullptr) ? dfs(root->left) : 0;
        int left = (root->right != nullptr) ? dfs(root->right) : 0;
        int depth = max(left, right) + 1;
        if (left + right > max_len) 
            max_len = left + right;
        return depth;
    }
    int diameterOfBinaryTree(TreeNode* root) {
        if (root == nullptr) return 0;
        dfs(root);
        return max_len;
    }
};