Problem
Given the root of a binary tree, return the preorder traversal of its nodes' values.
Example 1:
img
Input: root = [1,null,2,3]
Output: [1,2,3]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
img
Input: root = [1,2]
Output: [1,2]
Example 5:
img
Input: root = [1,null,2]
Output: [1,2]
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?
問題
給你二叉樹的根節點 root ，返回它節點值的 前序 遍歷。
示例 1：
img
輸入：root = [1,null,2,3]
輸出：[1,2,3]
示例 2：
輸入：root = []
輸出：[]
示例 3：
輸入：root = [1]
輸出：[1]
示例 4：
img
輸入：root = [1,2]
輸出：[1,2]
示例 5：
img
輸入：root = [1,null,2]
輸出：[1,2]
提示：
- 樹中節點數目在范圍 [0, 100] 內
- -100 <= Node.val <= 100
進階：遞歸算法很簡單，你可以通過迭代算法完成嗎？
思路
遞歸
根左右
先加入 root 節點的值，再遍歷左右子樹
Python3 代碼
# 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]:
# 遞歸
res = []
def dfs(root):
if not root:
return []
res.append(root.val)
dfs(root.left)
dfs(root.right)
dfs(root)
return res
迭代
使用棧來模擬
Python3 代碼
# 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]:
# 迭代
res = []
if not root:
return res
stack = []
node = root
while stack or node:
while node:
res.append(node.val)
stack.append(node)
# 前序遍歷
node = node.left
node = stack.pop()
node = node.right
return res
