LeetCode OJ--Path Sum II **

https://oj.leetcode.com/problems/path-sum-ii/

树的深搜，从根到叶子，并记录符合条件的路径。

注意参数的传递，是否需要使用引用。

#include <iostream>
#include <vector>
using namespace std;
struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 };


class Solution {
public:
    vector<vector<int> > ans;
    vector<vector<int> > pathSum(TreeNode *root, int sum) 
    {
        ans.clear();
        if(root == NULL)
            return ans;
        vector<int> ansPiece;
        hasPathSum(root,sum,ansPiece);
        return ans;
    }
    void hasPathSum(TreeNode *root, int sum, vector<int> ansPiece) {
        //null
        if(root == NULL )
            return ;

        //leaf node
        if(root->left == NULL && root->right == NULL && root->val == sum)
        {
            ansPiece.push_back(root->val);
            vector<int> _ansPiece = ansPiece;
            ans.push_back(_ansPiece);
            return ;
        }
        if(root->left == NULL && root->right == NULL)
            return ;

        //not leaf node
        if(root->left ||root->right)
         {
             ansPiece.push_back(root->val);

            if(root->left)
                hasPathSum(root->left, sum - root->val,ansPiece);

            if(root->right)
                hasPathSum(root->right, sum-root->val, ansPiece);  
         }
        return;
    }
};

int main()
{
    TreeNode *n1 = new TreeNode(1);
    TreeNode *n2 = new TreeNode(-2);
    TreeNode *n3 = new TreeNode(-3);
    TreeNode *n4 = new TreeNode(1);
    TreeNode *n5 = new TreeNode(3);
    TreeNode *n6 = new TreeNode(-2);
    TreeNode *n7 = new TreeNode(4);
    TreeNode *n8 = new TreeNode(4);
    n1->left = n2;
    n1->right = n3;
    n2->left = n4;
    n2->right = n5;
    n3->left = n6;
    n3->right = n8;
    n4->left = n7;
    class Solution myS;
    myS.pathSum(n1,2);
    return 0;
}