题目
给定一个二叉树的根节点 root ,返回它的 中序 遍历。
示例 1:
输入:root = [1,null,2,3]
输出:[1,3,2]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
示例 4:
输入:root = [1,2]
输出:[2,1]
示例 5:
输入:root = [1,null,2]
输出:[1,2]
提示:
- 树中节点数目在范围 [0, 100] 内
- -100 <= Node.val <= 100
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
代码
# 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: TreeNode) -> List[int]:
res = []
if root is not None:
stack = []
while stack or root is not None:
if root is not None:
stack.append(root)
root = root.left
else:
root = stack.pop()
res.append(root.val)
root = root.right
return res