题解 | #二叉树的前序遍历#

二叉树遍历

思路：

1. 没有返回，用一个self变量保存遍历的值。
2. 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