Given a binary tree, return the inorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]
1
2
/
3
Output: [1,3,2]
Follow up: Recursive solution is trivial, could you do it iteratively?
思路
这道题考察的是二叉树的中序遍历，中序遍历的解法有两种，一种是使用递归，一种是使用循环来解决。下面附上两种解法的代码。
解决代码
递归解决代码
1 # Definition for a binary tree node.
2 # class TreeNode(object):
3 # def __init__(self, x):
4 # self.val = x
5 # self.left = None
6 # self.right = None
7
8 class Solution(object):
9 def inorderTraversal(self, root):
10 """
11 :type root: TreeNode
12 :rtype: List[int]
13 """
14 if not root: # 空节点直接返回
15 return []
16 res =[]
17 self.indorder(root, res) # 中序遍历
18 return res
19
20 def indorder(self, root, res): # 中序遍历函数
21 if not root:
22 return
23
24 self.indorder(root.left, res)
25 res.append(root.val)
26 self.indorder(root.right, res)
循环解决代码(循环主要是使用栈来解决)
1 # Definition for a binary tree node.
2 # class TreeNode(object):
3 # def __init__(self, x):
4 # self.val = x
5 # self.left = None
6 # self.right = None
7
8 class Solution(object):
9 def inorderTraversal(self, root):
14 if not root:
15 return []
16 res = [] # 记录结果列表
17 stack = []
18 while stack or root: # 循环条件
19 if root: # 中序遍历，需要先将左边节点遍历完毕。
20 stack.append(root)
21 root = root.left
22 else: # 意味左节点为空，这时进行弹出。并遍历弹出节点的右子树。
23 root = stack.pop()
24 res.append(root.val)
25 root = root.right
26 return res