描述：
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree andsum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
思路：
这个题和上一题差不多，关键就是怎么处理存储的过成，看到大佬充分利用传值和传引用把这道题解决了，牛逼。
C++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:
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<vector<int>> res;
vector<int> tmp;
find(root,res,tmp,sum);
return res;
}
void find(TreeNode* root, vector<vector<int>> &res, vector<int> tmp,int sum)
{
if(root == NULL)
return;
tmp.push_back(root->val);
if(root->left==NULL && root->right ==NULL && root->val ==sum)
{
res.push_back(tmp);
}
find(root->left,res,tmp,sum-root->val);
find(root->right,res,tmp,sum-root->val);
}
};
Java代码：
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
ArrayList<ArrayList<Integer> > res = new ArrayList<>();
ArrayList<Integer> tmp = new ArrayList<>();
find(root,res,tmp,sum);
return res;
}
public void find(TreeNode root, ArrayList<ArrayList<Integer> > res,ArrayList<Integer> tmp,int sum){
if(root == null)
return ;
tmp.add(root.val);
if(root.left==null && root.right==null && root.val == sum){
res.add(tmp);
}
find(root.left, res, new ArrayList<Integer>(tmp), sum-root.val);
find(root.right, res, new ArrayList<Integer>(tmp), sum-root.val);
}
}
