思路:二叉树的前序遍历,存到res中返回。

Python:

#dfs
class
Solution: def levelOrder(self, root: TreeNode) -> List[int]: res=[] def dfs(root): if not root: return None res.append(root.val) dfs(root.left) dfs(root.right) dfs(root) return res
#bfs
class Solution:
    def levelOrder(self, root: TreeNode) -> List[int]:
        res,queue=[],[]
        if not root:
            return res
        queue.append(root)
        while queue:
            node=queue.pop(0)
            res.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        return res