112. Path Sum

递归：注意叶子节点的左右子树均为空

/**
 * Definition for a binary tree node.
 * 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) {
        
        return hasPath(root,sum);
            
    }
    bool hasPath(TreeNode *root,int sum)
    {
        if(root==NULL)
            return 0;
         if(root->left==NULL&&root->right==NULL&&(sum-root->val)==0)
             return 1;
         if(root->left==NULL&&root->right==NULL&&(sum-root->val)!=0)
             return 0;
        
        return hasPath(root->left,sum-root->val)|hasPath(root->right,sum-root->val);
    }
};