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 binary tree
 * 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 hasPathSum(root,sum,0);
    }

private:
   bool hasPathSum(TreeNode *root,int sum,int tempSum){
       if(root){
           tempSum+=root->val;
           if(NULL==root->left && NULL==root->right){
               if(sum==tempSum) return true;
           }
           bool left=false;
           bool right=false;
           if(root->left){
              left=hasPathSum(root->left,sum,tempSum);
           }
           if(root->right){
              right =hasPathSum(root->right,sum,tempSum);
           }
           return left || right;
       }else{
           return false;
       }
   }
};