[leetcode] Path sum路径之和

[b]要求[/b]
给定树，与路径和，判断是否存在从跟到叶子之和为给定值的路径。比如下图中，给定路径之和为22，存在路径<5,4,11,2>，因此返回true;否则返回false.
5
/ \
4   8
/   / \
11  13  4
/  \      \
7    2      5
[b]思路[/b]
递归，从跟到叶子判断，如果在叶子处剩下的给定值恰好为给定值，那么返回ture.
参考代码

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

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void addPath(TreeNode *root, int sum, vector<int> tmp, vector<vector<int> > &rev)
{
if (root != NULL && root->left == NULL && root->right == NULL && root->val == sum)
{
tmp.push_back(root->val);
rev.push_back(tmp);
tmp.pop_back();
}
if (root->left != NULL)
{
tmp.push_back(root->val);
addPath(root->left, sum - root->val, tmp, rev);
tmp.pop_back();
}
if (root->right != NULL)
{
tmp.push_back(root->val);
addPath(root->right, sum - root->val, tmp, rev);
tmp.pop_back();
}
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<vector<int> > rev;
if (root == NULL)
return rev;

vector<int> tmp;

addPath(root, sum, tmp, rev);
return rev;
}
};