-
Notifications
You must be signed in to change notification settings - Fork 0
/
94.BinaryTreeInorderTraversal(medium).cpp
76 lines (68 loc) · 1.47 KB
/
94.BinaryTreeInorderTraversal(medium).cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
/*
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> v;
if (root != NULL)
{ vector<int> v1,v2;
v1 = inorderTraversal(root->left);
for(int i = 0;i<v1.size();i++)
v.push_back(v1[i]);
v.push_back(root->val);
v2 = inorderTraversal(root->right);
for(int i = 0;i<v2.size();i++)
v.push_back(v2[i]);
}
return v;
}
};
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> v;
stack<TreeNode*> s;
if (root != NULL)
{
s.push(root);
while (!s.empty())
{
TreeNode* T = s.top();
s.pop();
if ( T->left == NULL)
{
v.push_back(T->val);
if(T->right != NULL)
s.push(T->right);
}
else
{
TreeNode* tmp = T->left;
T->left = NULL;
s.push(T);
s.push(tmp);
}
}
}
return v;
}
};