LeetCode_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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
queue<TreeNode *> myqueue;
if(root == NULL) return false;
myqueue.push(root) ;
bool flag = false ;
while(!myqueue.empty())
{
TreeNode * tp = myqueue.front();
myqueue.pop();

if(tp->left){
tp->left->val +=tp->val ;
myqueue.push(tp->left) ;
}
if(tp->right){
tp->right->val +=tp->val ;
myqueue.push(tp->right) ;

}
if(NULL == tp->left && NULL == tp->right )
{
if(tp->val == sum ) {
flag = true ;
break;
}
}

}

return flag ;
}
};

/**
* 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 DFS(TreeNode * root, int sum){
sum += root->val ;
if(NULL == root->left && root->right == NULL){
return sum == target;
}

if(root->left && DFS(root->left,sum)) return true;
if(root->right && DFS(root->right,sum) ) return true;
return false;
}
bool hasPathSum(TreeNode *root, int sum) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
target = sum;
if(root == NULL) return false;
return DFS(root, 0);
}

private :
int target;
};