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.

Note: A leaf is a node with no children.

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.

給一個二叉樹和一個值sum，判斷二叉樹是否有一條從根節點到葉子節點的路徑，使得路徑上節點值的和等於sum。

/**
 * 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) {
        if(root==NULL) return false;
        if(root->left==NULL&&root->right==NULL){
            if(sum==root->val) return true;
            else return false;
        }
        return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
       
    }
    
};