小知识，大挑战！本文正在参与“程序员必备小知识”创作活动。
描述
Given the root of a binary tree, return the inorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3]
Output: [1,3,2]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
Input: root = [1,2]
Output: [2,1]
Example 5:
Input: root = [1,null,2]
Output: [1,2]
Note:
The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100
解析
根据题意，就是给出了一个二叉树的根节点 root ，然后要求我们使用中序遍历这棵树，并将按照中序遍历顺序的节点按顺序放入列表中返回。这道题很简单，一般这种中序遍历的题，直接使用递归遍历每个节点就可以了。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def inorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
result = []
def dfs(root):
if not root:
return
dfs(root.left)
result.append(root.val)
dfs(root.right)
dfs(root)
return result
运行结果
Runtime: 20 ms, faster than 54.47% of Python online submissions for Binary Tree Inorder Traversal.
Memory Usage: 13.5 MB, less than 48.15% of Python online submissions for Binary Tree Inorder Traversal.
解析
当然了还可以使用栈来迭代遍历每个节点。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def inorderTraversal(self, root):
res, stack = [], []
while True:
while root:
stack.append(root)
root = root.left
if not stack:
return res
node = stack.pop()
res.append(node.val)
root = node.right
运行结果
Runtime: 46 ms, faster than 5.99% of Python online submissions for Binary Tree Inorder Traversal.
Memory Usage: 13.6 MB, less than 19.47% of Python online submissions for Binary Tree Inorder Traversal.
原题链接：leetcode.com/problems/bi…
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