【leetcode刷题】T124-路径总和 II

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]
]

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        res = []
        self.get_path_sum(root, [], res, sum)
        return res

    def get_path_sum(self, node, cur, res, sum):
        if not node:
            return
        if not node.left and not node.right:
            if node.val == sum:
                cur.append(node.val)
                res.append(cur)
            return 
        cur.append(node.val)
        self.get_path_sum(node.left, copy.copy(cur), res, sum-node.val)
        self.get_path_sum(node.right, copy.copy(cur), res, sum-node.val)

/**
 * Definition for a binary tree node.
 * 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> cur;
        get_path_sum(root, cur, res, sum);
        return res;
    }

    void get_path_sum(TreeNode* node, vector<int> cur, vector<vector<int>>& res, int sum){
        if(!node)
            return;
        if(!node->left && !node->right){
            if(node->val == sum){
                cur.push_back(node->val);
                res.push_back(cur);
            }
            return;
        }
        cur.push_back(node->val);
        get_path_sum(node->left, cur, res, sum-node->val);
        get_path_sum(node->right, cur, res, sum-node->val);
    }
};