算法和数据结构 路径总和

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        return dfs(root, sum);
    }
    bool dfs(TreeNode *t, int remain)
    {
        if(t == NULL) return false;
        else
        {
            // 叶子节点
            if(t->left == NULL && t->right == NULL)
            {
                return t->val == remain;
            }
            else
            {
                return dfs(t->left, remain - t->val) || dfs(t->right, remain - t->val);
            }
        }
    }
};