1 """
2 Given a binary tree, return the inorder traversal of its nodes' values.
3 Example:
4 Input: [1,null,2,3]
5 1
6 \
7 2
8 /
9 3
10 Output: [1,3,2]
11 """
12 """
13 中序遍历题
14 解法一递归
15 """
16 class TreeNode:
17 def __init__(self, x):
18 self.val = x
19 self.left = None
20 self.right = None
21
22 class Solution1:
23 def inorderTraversal(self, root):
24 if root == None:
25 return []
26 else:
27 List = []
28 self.order(root, List) #bug,刚开始没有把List传进去
29 return List
30 def order(self, root, List):
31 if root.left:
32 self.order(root.left, List)
33 List.append(root.val)
34 if root.right:
35 self.order(root.right, List)
36 """
37 解法二：非递归
38 栈 思路：
39 （1）树先一直向左走到叶节点并将沿途的结点入栈 ；
40 （2）然后向右走一步，重复第一步操作。
41 注意要将沿途的节点进行入栈
42 """
43 class TreeNode:
44 def __init__(self, x):
45 self.val = x
46 self.left = None
47 self.right = None
48 class Solution2(object):
49 def inorderTraversal(self, root):
50 """
51 :type root: TreeNode
52 :rtype: List[int]
53 """
54 if root == None:
55 return []
56 List = []
57 stack = []
58 while root or stack:
59 if root:
60 stack.append(root)
61 root = root.left
62 else:
63 root = stack.pop(-1)
64 List.append(root.val)
65 root = root.right
66 return List
67 if __name__ == '__main__':
68 S = Solution2()
69 l1 = TreeNode(1)
70 l2 = TreeNode(2)
71 l3 = TreeNode(3)
72 l4 = TreeNode(4)
73 l5 = TreeNode(5)
74 l6 = TreeNode(6)
75 l7 = TreeNode(7)
76 root = l1
77 l1.left = l2
78 l1.right = l3
79 l2.left = l4
80 l2.right = l5
81 l3.left = l6
82 l3.right = l7
83 print(S.inorderTraversal(root))
