【leetcode刷题】T133-二叉树的所有路径

示例:
输入:
   1
 /   \
2     3
 \
  5
输出: ["1->2->5", "1->3"]
解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

# 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 binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
        res = []
        self.get_path(root, [], res)
        return res

    def get_path(self, node, ls, res):
        ls2 = copy.copy(ls)
        ls2.append(node.val)
        if not node.left and not node.right:
            res.append("->".join(map(str, ls2)))
            return 
        if node.left:
            self.get_path(node.left, ls2, res)
        if node.right:
            self.get_path(node.right, ls2, res)

/**
 * 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<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if(!root)
            return res;
        get_path(root, "", res);
        return res;
    }

    void get_path(TreeNode* node, string str, vector<string>& res){
        string str2 = "";
        if(str == "")
            str2 = to_string(node->val);
        else
            str2 = str + "->" + to_string(node->val);
        if(!node->left && !node->right){
            res.push_back(str2);
            return;
        }
        if(node->left)
            get_path(node->left, str2, res);
        if(node->right)
            get_path(node->right, str2, res);
    }
};