【leetcode】path sum--easy

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / 
            4   8
           /   / 
          11  13  4
         /        
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

//二叉树为空时，全部返回false
//从根结点到叶子结点的和为sum，而不是到某一个结点处和为sum就行
//二叉树的值可以为负值
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 };
 
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if(root==NULL)
            return false;    
        else
        {
            int v=root->val;
            if(root->right==NULL&&root->left==NULL&&sum-v==0)
                return true;
            else if(root->left!=NULL&&hasPathSum(root->left,sum-v)==true)
                return true;
            else if(root->right!=NULL&&hasPathSum(root->right,sum-v)==true)
                return true;
            else
                return false;
        }
    }
};