257. Binary Tree Paths (树的路径)

 

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

 

   1
 /   \
2     3
 \
  5

 

All root-to-leaf paths are:

["1->2->5", "1->3"]

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def __init__(self):
        self.res = []
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        def dfs(root,path):
            if root == None:
                return
            if root and root.left == None and root.right == None:
                path.append(str(root.val))
                self.res.append('->'.join(path))
            dfs(root.left,path+[str(root.val)])
            dfs(root.right,path+[str(root.val)])
        dfs(root,[])
        return self.res
            

 

class Solution {
public:
    vector<vector<int>> res;
    void backtrack(TreeNode* root, vector<int>& path) {  
        if (root == nullptr) return;
        if(root->left == nullptr && root->right == nullptr) {
            path.push_back(root->val);
            res.push_back(path);
            path.pop_back();
            return;
        }
        path.push_back(root->val);
        backtrack(root->left,path);
        backtrack(root->right,path);
        path.pop_back();
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> final_res;
        vector<int> path;
        backtrack(root,path);
        for(auto vec: res){
            string path_str = "";
            for(auto int i = 0;i < vec.size(); ++i){
                path_str.append(to_string(vec[i]));
                if (i != vec.size() - 1) path_str.append("->");
            }
            final_res.emplace_back(path_str);
        }
        return final_res;
    }
};