"Leetcode" Binary Tree Preorder traversal

Given a binary tree, return the preorder traversal of its nodes ' values.

For example:

Given binary Tree {1,#,2,3},

1
\
2
/
3
return [+].

Can be recursive, proceed. You can also use iterations.

C++:

1 classSolution {2  Public:3     voidPreorder (TreeNode * root, vector<int> &path)4     {   5         if(Root)6         {7Path.push_back (root->val);8Preorder (root->Left,path);9Preorder (root->Right,path);Ten         } One     } Avector<int> Preordertraversal (TreeNode *root) { -vector<int>path; -         if(Root==null)returnpath; the preorder (root,path); -         returnpath; -     } -};

1 classsolution{2  Public:3vector<int> Preordertraversal (TreeNode *root) {4vector<int>path;5Stack<treenode*>Stk;6          while(root!=null| |!stk.empty ())7         {8             if(root!=NULL)9             {Ten                  while(Root) One                 { APath.path_back (root->val); - Stk.push (root); -Root=root->Left ; the                 } -}Else{ -Root=stk.top ()Right ; - Stk.pop (); +             } -         } +         returnpath; A     } at};

Python:

1 #Definition for a binary tree node2 #class TreeNode:3 #def __init__ (self, x):4 #self.val = x5 #self.left = None6 #self.right = None7 8 classSolution:9     #@param root, a tree nodeTen     #@return A list of integers One     defpreordertraversal (self, root): A         ifroot==None: -             return [] -         if(Root): the             return[Root.val]+self.preordertraversal (Root.left) +self.preordertraversal (root.right) -              -             

"Leetcode" Binary Tree Preorder traversal