94. 二叉树的中序遍历
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?
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-inorder-traversal
Link:https://leetcode.com/problems/binary-tree-inorder-traversal/
递归
O(N)
中序遍历,就是左->中->右
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
res = []
self.traverse(root, res)
return res
def traverse(self, node: TreeNode, res: List[int]):
if node is None:
return
self.traverse(node.left, res)
res.append(node.val)
self.traverse(node.right, res)
非递归
O(N)
这个没法解释,只能自己尝试着推导一下
首先,要输出最左边的左节点, 那之前的路径用stack保存下来, 用来回退到中节点
结尾的时候,当前值cur要变化,那把它置为右节点
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
res = []
stack = []
cur = root
while cur is not None or len(stack) > 0:
while cur is not None and cur.left is not None:
stack.append(cur)
cur = cur.left
if cur is None:
cur = stack.pop()
res.append(cur.val)
cur = cur.right
return res
多入栈一次,少了一些判断
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
res = []
stack = []
cur = root
while cur is not None or len(stack) > 0:
while cur is not None:
stack.append(cur)
cur = cur.left
cur = stack.pop()
res.append(cur.val)
cur = cur.right
return res
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