题解 | 二叉树中和为某一值的路径(一)

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
bool ok = false;
#include <cstddef>
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 
     * @param sum int整型 
     * @return bool布尔型
     */
    bool hasPathSum(TreeNode* root, int sum, int res = 0) {
        // write code here
        if (not root) return false;

        res += root->val;
        if (not root->left and not root->right and res == sum) return true;

        return hasPathSum(root->left, sum, res) or hasPathSum(root->right, sum, res);
    }
};