题解 | #二叉树的前序遍历#
二叉树的前序遍历
https://www.nowcoder.com/practice/5e2135f4d2b14eb8a5b06fab4c938635
二叉树遍历
思路：
- 没有返回，用一个self变量保存遍历的值。
- 用递归
前序遍历 - 根左右
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# @param root TreeNode类
# @return int整型一维数组
class Solution:
res = []
def preorderTraversal(self , root: TreeNode) -> List[int]:
# 根左右
if root == None:
return
self.res.append(root.val)
self.preorderTraversal(root=root.left)
self.preorderTraversal(root=root.right)
return self.res
中序遍历 - 左根右
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# @param root TreeNode类
# @return int整型一维数组
# 左根右
class Solution:
res = []
def inorderTraversal(self , root: TreeNode) -> List[int]:
# write code here
if root == None:
return
self.inorderTraversal(root.left)
self.res.append(root.val)
self.inorderTraversal(root.right)
return self.res
后序遍历 - 左右根
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# @param root TreeNode类
# @return int整型一维数组
# 左右根
class Solution:
res = []
def postorderTraversal(self , root: TreeNode) -> List[int]:
# write code here
if root == None:
return
self.postorderTraversal(root.left)
self.postorderTraversal(root.right)
self.res.append(root.val)
return self.res
