分析路径走向的三种可能即可

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型
     */
    int result = 0;
    int dfs(TreeNode* root){
        if(root == NULL)
            return 0;
        int left = dfs(root->left);
        int right = dfs(root->right);
        result = max(result,max(left,right) + 1);
        result = max(result,left + right + 1);
        return (left > right ? left : right) + 1;
    }
    int diameterOfBinaryTree(TreeNode* root) {
        // write code here
        if(root == NULL)
            return 0;
        dfs(root);
        return result - 1;
    }
};