leetcode_144. 二叉树的前序遍历

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。

 

示例 1：


输入：root = [1,null,2,3]
输出：[1,2,3]
示例 2：

输入：root = []
输出：[]
示例 3：

输入：root = [1]
输出：[1]
示例 4：


输入：root = [1,2]
输出：[1,2]
示例 5：


输入：root = [1,null,2]
输出：[1,2]
 

提示：

树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100
 

进阶：递归算法很简单，你可以通过迭代算法完成吗？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-preorder-traversal
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

# 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 preorderTraversal(self, root: TreeNode) -> List[int]:
        if not root :return []
        return [root.val]+self.preorderTraversal(root.left)+self.preorderTraversal(root.right)

# 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 preorderTraversal(self, root: TreeNode) -> List[int]:
        def preOrder(r:TreeNode)->None:
            if not r:return
            ls.append(r.val)
            preOrder(r.left)
            preOrder(r.right)
                 
        ls=[]
        preOrder(root)
        return ls

# 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 preorderTraversal(self, root: TreeNode) -> List[int]:
        if not root :return []
        ls=[]
        stack=[]
        cur=root
        
        while(stack or cur):
            while cur:
                ls.append(cur.val)
                stack.append(cur)
                cur=cur.left
            cur=stack.pop()            
            cur=cur.right
        return ls