path Sum

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.

/**
 * 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;
        }
        //leaf node ,no left and right recurision end condition
        if(root->left==NULL && root->right==NULL){
            return root->val==sum;
        }
      //left right exist one,return true
        bool left=hasPathSum(root->left,sum-root->val);
        bool right=hasPathSum(root->right,sum-root->val);
            
        return left || right;

    }
};