算法: 144. 二叉树前序遍历Binary Tree Preorder Traversal

Posted 2022-12-15 AI架构师易筋

144. Binary Tree Preorder Traversal

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

Example 1:

Input: root = [1,null,2,3]
Output: [1,2,3]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

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

1. 递归求解 – 计算机思维

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

2. 遍历求解 – 人类思维

# 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: Optional[TreeNode]) -> List[int]:
        res = []
        stack = [root]
        while stack:
            node = stack.pop()
            if not node: continue
            res.append(node.val)
            stack.append(node.right)
            stack.append(node.left)

        return res