给定一个二叉树和一个目标和,找到所有从根节点到叶子节点路径总和等于给定目标和的路径。
说明: 叶子节点是指没有子节点的节点。
示例:
给定如下二叉树,以及目标和 sum = 22,
5 / 4 8 / / 11 13 4 / / 7 2 5 1
返回:
[ [5,4,11,2], [5,8,4,5] ]
class Solution {
public:
vector<vector<int> > res;
vector<vector<int> > pathSum(TreeNode* root, int sum)
{
if(root == NULL)
{
return res;
}
vector<int> v;
DFS(root, v, sum);
return res;
}
void DFS(TreeNode* root, vector<int> &v, int sum)
{
if(root == NULL)
return;
if(root ->left == NULL && root ->right == NULL)
{
if(sum == root ->val)
{
v.push_back(root ->val);
res.push_back(v);
v.pop_back();
}
return;
}
v.push_back(root ->val);
if(root ->left)
DFS(root ->left, v, sum - root ->val);
if(root ->right)
DFS(root ->right, v, sum - root ->val);
v.pop_back();
}
};