144. 二叉树前序遍历

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

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

 

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

 

 

 

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        cur = root
        stack = []
        while cur or stack:
            while cur :
                res.append(cur.val)
                stack.append(cur)
                cur = cur.left
            t = stack.pop()
            cur = t.right
        return res
        

 

 

 

 

 

class Solution {
    private List<Integer> res = new ArrayList<Integer>();
    public List<Integer> preorderTraversal(TreeNode root) {
        help(root);
        return res;
    }
    private void help(TreeNode root){
        if(root == null) return ;
        res.add(root.val);
        help(root.left);
        help(root.right);
    }
}