判断是否存在二叉树中和为某一值的路径

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    bool hasPathSum(TreeNode* root, int sum) 
    {
        if(root == nullptr) return false;
        else if(root -> val == sum && root -> left == nullptr && root -> right == nullptr)
        {
            return true;
        }
        return hasPathSum(root -> left, sum - root -> val) || hasPathSum(root -> right, sum - root -> val);
    }
    
};