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.

DFS:

bool findPathSum(TreeNode* root, vector<int> &path, int sum)
{
path.push_back(root->val);
if(root->left == NULL && root->right == NULL){
vector<int>::iterator it = path.begin();
int tmpsum = 0;
for(; it != path.end(); ++it)
tmpsum += *it;
path.pop_back();
if(tmpsum == sum)
return true;
else
return false;
}
bool flag = false;
if(root->left)
flag = findPathSum(root->left,path,sum);
if(!flag && root->right)
flag = findPathSum(root->right,path,sum);
path.pop_back();
return flag;
}
bool hasPathSum(TreeNode *root, int sum) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root == NULL)
return false;
vector<int> path;
return findPathSum(root, path,sum);
}

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