【LeetCode每天一题】Binary Tree Preorder Traversal(前序遍历)

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

Example:

Input: [1, null, 1,2,3 ]
   1
         2
    /
   3

Output: [1,2,3]

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

思路

解决代码

 1 # Definition for a binary tree node.
 2 # class TreeNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7
 8 class Solution(object):
 9     def preorderTraversal(self, root):
10         """
11         :type root: TreeNode
12         :rtype: List[int]
13         """
14         if not root:
15             return []
16
17         res = []          # 结果集
18         stack = [root]      # 辅助栈
19         while stack:          # 循环条件
20             tem = stack.pop()     # 弹出最上层的节点
21             if tem.right:          # 先判断当前节点的右节点
22                 stack.append(tem.right)
23             if tem.left:             # 在判断当前节点的左节点
24                 stack.append(tem.left)
25             res.append(tem.val)          # 将节点值添加进结果集中
26         return res             # 返回结果集

递归解决代码

 1 # Definition for a binary tree node.
 2 # class TreeNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7
 8 class Solution(object):
 9     def preorderTraversal(self, root):
10         """
11         :type root: TreeNode
12         :rtype: List[int]
13         """
14         if not root:      # 为空直接返回
15             return
16         res = []
17         self.tarversal(root, res)  # 开始递归
18         return res
19
20     def tarversal(self, root, res):
21         if not root:        # 当前为空节点直接返回
22             return
23         res.append(root.val)      # 将当前节点的值添加进其中
24         self.tarversal(root.left, res)      # 遍历左节点
25         self.tarversal(root.right, res)     # 遍历右节点

原文地址：https://www.cnblogs.com/GoodRnne/p/10963509.html