python LeetCode 刷题记录 94

题目

给定一个二叉树的根节点 root ，返回 它的 中序 遍历

代码

递归

# 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 inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:if not root:return []ls = []ls += self.inorderTraversal(root.left)ls.append(root.val)ls += self.inorderTraversal(root.right)return ls

迭代

# 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 inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:if not root:return []ls = []stack = []current = rootwhile current or stack:while current:stack.append(current)current = current.left# 此循环出去时current是Nonecurrent = stack.pop() # 此时current是最左侧节点ls.append(current.val)current = current.right # 判断其右侧节点，如果有则，则重复上面的第二层while循环，否则会继续pop节点出来return ls