给定一个二叉树,返回它的中序 遍历。
示例:
输入: [1,null,2,3]
1
\
2
/
3
输出: [1,3,2]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
递归
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def inorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if not root:
return []
return self.inorderTraversal(root.left)+[root.val]+self.inorderTraversal(root.right)
循环
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def inorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
l = []
node_l = []
left = root
while True:
if left:
node_l.append(left)
left = left.left
elif node_l:
node = node_l.pop()
l.append(node.val)
left = node.right
else:
break
return l
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