-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathBinaryTreePostorderTraversal.cpp
More file actions
69 lines (60 loc) · 1.62 KB
/
BinaryTreePostorderTraversal.cpp
File metadata and controls
69 lines (60 loc) · 1.62 KB
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
/**
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?
[Analysis]
The idea is to use two stacks.
One stack store the sequence of pre-order sequence
any time when a node is poped from the pre-order stack
push it into the post-order stack
the last step is to pop every element from the post-order stack and put them into
result set
*/
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> res;
if (!root) return res;
stack<TreeNode*> s1;
stack<TreeNode*> s2;
s1.push(root);
// Run while first stack is not empty
while (!s1.empty())
{
// Pop an item from s1 and push it to s2
TreeNode* top = s1.top();
s1.pop();
s2.push(top);
// Push left and right children of removed item to s1
//note that for pre-order, first push right, then left
//here we do the opposite
if (top->left)
s1.push(top->left);
if (top->right)
s1.push(top->right);
}
// Print all elements of second stack
while (!s2.empty())
{
TreeNode* top = s2.top();
res.push_back(top->val);
s2.pop();
}
}
};