-
Notifications
You must be signed in to change notification settings - Fork 2
/
binary-tree-postorder-traversal.cpp
56 lines (49 loc) · 1.2 KB
/
binary-tree-postorder-traversal.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
题目描述
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},
1
\
2
/
3
return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
时间复杂度O(n),空间复杂度O(n)
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
stack<const TreeNode *> s;
const TreeNode *p = root, *q = nullptr;
do{
while(p != nullptr){
s.push(p);
p = p->left;
}
q = nullptr;
while(!s.empty()){
p = s.top();
s.pop();
if(p->right == q){
result.push_back(p->val);
q = p;
}else{
s.push(p);
p = p->right;
break;
}
}
}while(!s.empty());
return result;
}
};